- Look for an equation that can be written as a product of a function of \(x\) and a function of \(y\). - After choosing an equation, move the y-dependent factor to the side with \(dy\).

1. Choice B has the product form \(y'=xy\). 2. For \(y\ne0\), it separates as \(\frac{1}{y}\,dy=x\,dx\).

B. For \(y\ne0\), \(\frac{1}{y}\,dy=x\,dx\).

- Put every factor involving \(y\) on the side with \(dy\). - Put every factor involving \(x\) on the side with \(dx\).

1. Multiply by \(1+y^2\) to place the y-dependent factor with \(dy\). 2. The separated form is \((1+y^2)\,dy=x^2\,dx\).

\((1+y^2)\,dy=x^2\,dx\)

- Try moving all the \(y\)-dependent terms to the side with \(dy\). - Does either equation still mix \(x\) and \(y\) on one side after rearranging?

1. The equation \(y'=xy\) is already a product of an x-only factor and a y-only factor. 2. The sum \(x+y\) does not have that product form.

\(y'=xy\) is separable because its right side is the product of \(x\) and \(y\).

- Write the prime as a differential quotient. - Move the factor containing \(y\) to the side with \(dy\).

1. Rewrite the derivative as \(\frac{dy}{dx}=2x(1+y^2)\). 2. Since \(1+y^2>0\), divide by it to obtain \(\frac{1}{1+y^2}\,dy=2x\,dx\).

\(\frac{1}{1+y^2}\,dy=2x\,dx\).

- Integrate the two sides separately. - Which logarithmic function has derivative \(1/y\) when \(y\) is nonzero? - One arbitrary constant is enough after integration.

1. An antiderivative of \(1/y\) is \(\ln|y|\). 2. Integrating both sides gives \(\ln|y|=3x+C\).

\(\ln|y|=3x+C\).

- What is the derivative of a constant function? - Look for values that make the factor you intend to divide by zero.

1. Constant solutions have \(y'=0\). 2. Solving \(y(1-y)=0\) gives \(y=0\) or \(y=1\); division by that product would exclude both.

\(y=0\) and \(y=1\).

- For nonzero \(y\), move \(y\) to the differential side to obtain a logarithmic integral. - Integrate the \(x\)-side using the power rule, then exponentiate the constant into a multiplicative constant. - Check \(y=0\) separately because dividing by \(y\) can remove it.

1. For \(y\ne0\), separate the variables: \(\frac{1}{y}\,dy=x\,dx\). 2. Antidifferentiate: \(\ln|y|=\frac{x^2}{2}+C\). 3. Exponentiate and absorb the sign and positive factor into one nonzero constant: \(y=Ce^{x^2/2}\), where \(C\ne0\). 4. The divided-out case \(y=0\) is also a solution, so allowing \(C=0\) gives the full family.

\(y=Ce^{x^2/2}\), where \(C\in\mathbb R\)

- Separate the equation as a logarithmic \(y\)-integral and a polynomial \(x\)-integral. - The antiderivative of \(3x^2\) determines the exponent in the explicit family. - Restore the zero solution after dividing by \(y\), and check whether the final formula has any finite-domain restriction.

1. For \(y\ne0\), separate the variables: \(\frac{1}{y}\,dy=3x^2\,dx\). 2. Antidifferentiate: \(\ln|y|=x^3+C\). 3. Exponentiate to obtain \(y=Ce^{x^3}\) for \(C\ne0\). 4. The constant solution \(y=0\) is included by allowing \(C=0\). Each solution is defined for all real \(x\).

\(y=Ce^{x^3}\), where \(C\in\mathbb R\); each solution is defined for all \(x\in\mathbb R\)

- How can you get the terms involving \(y\) onto the same side as \(dy\)? - The antiderivative of \(e^x\) remains exponential. - Could dividing by an expression involving \(y\) cause you to miss a solution? Check the original equation.

1. For \(y\ne0\), separate the variables: \(\frac{1}{y}\,dy=e^x\,dx\). 2. Antidifferentiate: \(\ln|y|=e^x+C\). 3. Exponentiate to obtain \(y=Ce^{e^x}\) for \(C\ne0\). 4. The divided-out solution \(y=0\) is included by allowing \(C=0\).

\(y=Ce^{e^x}\), where \(C\in\mathbb R\)

- Separate as \(dy/y=\cos x\,dx\) on nonzero solution branches. - Use the antiderivative of cosine before exponentiating. - Verify that the multiplicative-constant form also includes the equilibrium solution \(y=0\).

1. For \(y\ne0\), separate the variables: \(\frac{1}{y}\,dy=\cos x\,dx\). 2. Antidifferentiate: \(\ln|y|=\sin x+C\). 3. Exponentiate to obtain \(y=Ce^{\sin x}\) for \(C\ne0\). 4. The divided-out solution \(y=0\) is included by allowing \(C=0\).

\(y=Ce^{\sin x}\), where \(C\in\mathbb R\)

- Record first that the differential equation is undefined at \(x=0\). - On a nonzero branch, separate into \(dy/y=dx/x\) and use logarithms with absolute values. - Check \(y=0\) separately and keep each solution on an interval entirely to one side of \(x=0\).

1. On an interval where \(x\ne0\) and for \(y\ne0\), separate the variables: \(\frac{1}{y}\,dy=\frac{1}{x}\,dx\). 2. Antidifferentiate: \(\ln|y|=\ln|x|+C\). 3. Exponentiate and absorb the signs into one constant to obtain \(y=Cx\) for \(C\ne0\). 4. The divided-out solution \(y=0\) is included by allowing \(C=0\). Solutions are defined on intervals that do not contain \(x=0\).

\(y=Cx\), where \(C\in\mathbb R\), on any interval not containing \(x=0\)

- Divide by \(y\) only after noting the possible equilibrium solution. - Integrate \(x+1\) term by term to form the exponent. - Absorb the exponential of the integration constant into one real multiplicative constant.

1. For \(y\ne0\), separate the variables: \(\frac{1}{y}\,dy=(x+1)\,dx\). 2. Antidifferentiate: \(\ln|y|=\frac{x^2}{2}+x+C\). 3. Exponentiate to obtain \(y=Ce^{x^2/2+x}\) for \(C\ne0\). 4. The divided-out solution \(y=0\) is included by allowing \(C=0\).

\(y=Ce^{x^2/2+x}\), where \(C\in\mathbb R\)

- Before dividing by a factor containing \(y\), consider whether it can be zero. - Use a substitution or reverse chain rule to integrate \(\sin(2x)\). - Check whether any constant solutions were excluded while separating variables.

1. For \(y\ne0\), separate the variables: \(\frac{1}{y}\,dy=\sin(2x)\,dx\). 2. Antidifferentiate: \(\ln|y|=-\frac{1}{2}\cos(2x)+C\). 3. Exponentiate to obtain \(y=Ce^{-\frac{1}{2}\cos(2x)}\) for \(C\ne0\). 4. The divided-out solution \(y=0\) is included by allowing \(C=0\). The differential equation and all solutions are defined for every real \(x\).

\(y=Ce^{-\frac{1}{2}\cos(2x)}\), where \(C\in\mathbb R\); the equation is defined for all \(x\in\mathbb R\)

- For \(y\ne0\), separate using \(dy/y=dx/(1+x^2)\). - Recognize the inverse-tangent antiderivative on the \(x\)-side. - Include \(y=0\) and note that \(1+x^2\) never vanishes for real \(x\).

1. For \(y\ne0\), separate the variables: \(\frac{1}{y}\,dy=\frac{1}{1+x^2}\,dx\). 2. Antidifferentiate: \(\ln|y|=\arctan x+C\). 3. Exponentiate to obtain \(y=Ce^{\arctan x}\) for \(C\ne0\). 4. The divided-out equilibrium solution \(y=0\) is included by allowing \(C=0\).

\(y=Ce^{\arctan x}\), where \(C\in\mathbb R\)

- Multiply by \(y\,dx\) so the separated relation has \(y\,dy\) on one side. - After integration, solve the square relation as separate positive and negative branches when needed. - Because the original quotient contains \(y\), a valid branch cannot include a point where \(y=0\).

1. Because the differential equation is defined only when \(y\ne0\), work on an interval where \(y\) does not vanish. 2. Separate the variables: \(y\,dy=2x\,dx\). 3. Antidifferentiate: \(\frac{y^2}{2}=x^2+C\), or \(y^2=2x^2+C\). 4. Each solution branch must remain in a region where \(2x^2+C>0\), with one consistent choice of sign for \(y\).

\(y^2=2x^2+C\), on intervals where \(2x^2+C>0\) and \(y\ne0\)

- Separate by multiplying the equation by \(y\,dx\). - Integrate \(x^2\) and \(y\) with the power rule, then keep the result in an implicit square form if convenient. - Test \(y=0\) in the original equation; do not label an excluded value as a constant solution.

1. The differential equation is defined only when \(y\ne0\). 2. Separate the variables: \(y\,dy=x^2\,dx\). 3. Antidifferentiate: \(\frac{y^2}{2}=\frac{x^3}{3}+C\), or \(y^2=\frac{2}{3}x^3+C\). 4. Solution branches lie on intervals where \(\frac{2}{3}x^3+C>0\). There is no constant solution because \(y=0\) is outside the differential equation's domain and no nonzero constant can satisfy the equation for all \(x\).

\(y^2=\frac{2}{3}x^3+C\), on intervals where \(\frac{2}{3}x^3+C>0\); there is no constant solution

- Multiply by \(y\,dx\) to obtain a product suitable for direct integration. - Use the antiderivative of \(\sin x\) and retain an arbitrary constant. - Check the original quotient at \(y=0\) before claiming any constant solution.

1. The differential equation is defined only when \(y\ne0\). 2. Separate the variables: \(y\,dy=\sin x\,dx\). 3. Antidifferentiate: \(\frac{y^2}{2}=-\cos x+C\), or \(y^2=-2\cos x+C\). 4. Solution branches lie on intervals where \(C-2\cos x>0\). No constant solution exists because \(y=0\) is not in the domain and a nonzero constant would require \(\sin x=0\) for all \(x\).

\(y^2=-2\cos x+C\), on intervals where \(C-2\cos x>0\); there is no constant solution

- Multiply by \(1+y^2\) to put all \(y\)-terms with \(dy\). - Integrate the polynomial in \(y\) and the linear function of \(x\). - An implicit form is natural here; check whether the left side is one-to-one before forcing an explicit formula.

1. Separate the variables: \((1+y^2)\,dy=x\,dx\). 2. Antidifferentiate: \(y+\frac{y^3}{3}=\frac{x^2}{2}+C\). 3. Absorb additive constants into one constant and write the solution as \(y+\frac{y^3}{3}=\frac{x^2}{2}+C\).

\(y+\frac{y^3}{3}=\frac{x^2}{2}+C\)

- Move \(y^2\) to the differential side to obtain \(y^2\,dy=\cos x\,dx\). - Integrate with the power rule and the sine antiderivative. - The original equation is undefined at \(y=0\), so restrict any branch that reaches that value.

1. The differential equation is defined only when \(y\ne0\). 2. Separate the variables: \(y^2\,dy=\cos x\,dx\). 3. Antidifferentiate: \(\frac{y^3}{3}=\sin x+C\), or \(y^3=3\sin x+C\). 4. Each solution branch is restricted to an interval on which \(3\sin x+C\ne0\).

\(y^3=3\sin x+C\), on intervals where \(3\sin x+C\ne0\)

- Before dividing by \(y-2\), test whether \(y=2\) is an equilibrium solution. - For nonconstant branches, integrate \(dy/(y-2)=x\,dx\). - Exponentiate the logarithmic relation and let the multiplicative constant represent both signs.

1. For \(y\ne2\), separate the variables: \(\frac{1}{y-2}\,dy=x\,dx\). 2. Antidifferentiate: \(\ln|y-2|=\frac{x^2}{2}+C\). 3. Exponentiate to obtain \(y=2+Ce^{x^2/2}\) for \(C\ne0\). 4. The divided-out equilibrium solution \(y=2\) is included by allowing \(C=0\). Every solution is defined for all real \(x\).

\(y=2+Ce^{x^2/2}\), where \(C\in\mathbb R\); each solution is defined for all \(x\in\mathbb R\)

- Check \(y=-3\) separately before dividing by \(y+3\). - Integrate \(dy/(y+3)=\cos x\,dx\). - After exponentiating, include the equilibrium through the zero value of the multiplicative constant.

1. For \(y\ne-3\), separate the variables: \(\frac{1}{y+3}\,dy=\cos x\,dx\). 2. Antidifferentiate: \(\ln|y+3|=\sin x+C\). 3. Exponentiate to obtain \(y=-3+Ce^{\sin x}\) for \(C\ne0\). 4. The divided-out equilibrium solution \(y=-3\) is included by allowing \(C=0\).

\(y=-3+Ce^{\sin x}\), where \(C\in\mathbb R\)

- Multiply by \(y^2\,dx\) to obtain a power integral in \(y\). - Integrate both sides and retain the implicit cubic relation if it is clearer. - Because the original equation divides by \(y^2\), a solution branch cannot pass through \(y=0\).

1. The differential equation is defined only when \(y\ne0\). 2. Separate the variables: \(y^2\,dy=x\,dx\). 3. Antidifferentiate: \(\frac{y^3}{3}=\frac{x^2}{2}+C\), or \(y^3=\frac{3}{2}x^2+C\). 4. Each solution branch is restricted to an interval on which \(\frac{3}{2}x^2+C\ne0\).

\(y^3=\frac{3}{2}x^2+C\), on intervals where \(\frac{3}{2}x^2+C\ne0\)

- Move \((y-4)^2\) to the side with \(dy\). - Integrate the shifted power using \(u=y-4\) or the reverse chain rule. - The original equation is undefined at \(y=4\); check that this value is not mistakenly added as a constant solution.

1. The differential equation is defined only when \(y\ne4\). 2. Separate the variables: \((y-4)^2\,dy=x\,dx\). 3. Antidifferentiate: \(\frac{(y-4)^3}{3}=\frac{x^2}{2}+C\), or \((y-4)^3=\frac{3}{2}x^2+C\). 4. Solution branches must remain where \(\frac{3}{2}x^2+C\ne0\). The excluded value \(y=4\) is not a solution.

\((y-4)^3=\frac{3}{2}x^2+C\), on intervals where \(\frac{3}{2}x^2+C\ne0\); \(y=4\) is not a solution

- Which factor belongs on the same side as \(dy\) when the variables are separated? - Integrate both sides term by term to obtain an implicit quadratic relation in \(y\). - Check which values would make the original denominator zero before stating valid branches.

1. The differential equation is defined only when \(y\ne-\frac{1}{2}\). 2. Separate the variables: \((2y+1)\,dy=(x+1)\,dx\). 3. Antidifferentiate: \(y^2+y=\frac{x^2}{2}+x+C\). 4. Each solution branch must remain on an interval where \(y\ne-\frac{1}{2}\); the excluded value is not a solution.

\(y^2+y=\frac{x^2}{2}+x+C\), on branches where \(y\ne-\frac{1}{2}\)

- Move \(3y^2\) to the differential side before integrating. - Use the power rule to obtain a cubic relation in \(y\). - The original quotient excludes \(y=0\), so restrict a branch if the implicit relation reaches that value.

1. The differential equation is undefined when \(y=0\). 2. Separate the variables: \(3y^2\,dy=2x\,dx\). 3. Antidifferentiate: \(y^3=x^2+C\). 4. Equivalently, \(y=\sqrt[3]{x^2+C}\), restricted to intervals where \(x^2+C\ne0\).

\(y^3=x^2+C\), on intervals where \(x^2+C\ne0\); the differential equation is undefined at \(y=0\)

- Multiply by \(y^2+1\) to separate the variables without dividing by a potentially zero real factor. - Integrate the polynomial expressions on both sides term by term. - A constant solution would require the numerator \(x^2+1\) to vanish for all \(x\); test that possibility directly.

1. Since \(y^2+1>0\), separate the variables without excluding any real value of \(y\): \((y^2+1)\,dy=(x^2+1)\,dx\). 2. Antidifferentiate: \(\frac{y^3}{3}+y=\frac{x^3}{3}+x+C\). 3. There is no equilibrium solution because \(x^2+1\) is never \(0\), so a constant function cannot satisfy the equation.

\(\frac{y^3}{3}+y=\frac{x^3}{3}+x+C\); there is no equilibrium solution

- After separation, the \(x\)-side is \(4x\,dx\), not a constant multiple of \(x^2\,dx\). - Apply the power rule to \(4x\) and compare the derivative of the proposed exponent with the original coefficient. - Check \(y=0\) separately because the logarithmic step assumes \(y\ne0\).

1. Rewrite or inspect the equation so that each variable is confined to one side. 2. Since \(\int 4x\,dx=2x^2\), the corrected result is \(\ln|y|=2x^2+C\), so \(y=Ce^{2x^2}\).

\(\int 4x\,dx=2x^2\), so \(\ln|y|=2x^2+C\) and \(y=Ce^{2x^2}\).

- Which polynomial terms would have derivatives matching the two sides? - How does the power rule change the coefficients when you antidifferentiate? - Combine the two integration constants into a single constant on one side.

1. The variables are already separated, so integrate each side directly. 2. Integration gives \(y^3=x^2+x+C\).

\(y^3=x^2+x+C\)

- Move \(1-y^2\) to the same side as \(dy\). - Integrate the polynomial in \(y\) without trying to solve the resulting cubic. - Check where the original right side is undefined before stating valid solution branches.

1. The differential equation is defined only when \(y\ne-1\) and \(y\ne1\). 2. Separate the variables: \((1-y^2)\,dy=2x\,dx\). 3. Integrate to obtain \(y-\frac{y^3}{3}=x^2+C\). 4. Each differentiable solution branch must remain in one of the regions \(y<-1\), \(-1<y<1\), or \(y>1\).

\(y-\frac{y^3}{3}=x^2+C\), on branches that do not cross \(y=-1\) or \(y=1\)

- Move the entire denominator involving \(y\) to the differential side with \(dy\). - Integrate the polynomial terms separately. - Distinguish a value lost by division from a value where the original equation is undefined.

1. Multiply by \((y^2-1)\,dx\) to obtain \((y^2-1)\,dy=x^2\,dx\). 2. Integrating gives \(\frac{y^3}{3}-y=\frac{x^3}{3}+C\). 3. Because the original right side is undefined at \(y=1\) and \(y=-1\), every valid branch must avoid both levels. 4. Neither excluded level is an equilibrium solution because neither can be substituted into the original differential equation.

Separated equation: \((y^2-1)\,dy=x^2\,dx\) Integrated relation: \(\frac{y^3}{3}-y=\frac{x^3}{3}+C\) Restrictions: \(y\ne1\) and \(y\ne-1\) throughout a solution branch. Neither excluded level is an equilibrium because the differential equation is undefined there.

- Move the dependent-variable denominator to the side with \(dy\). - An implicit relation is the natural endpoint after integration. - Preserve the value excluded by the original denominator.

1. Rearrange as \((y^3+1)\,dy=(x^2+1)\,dx\). 2. Integration gives \(\frac{y^4}{4}+y=\frac{x^3}{3}+x+C\). 3. Valid branches must avoid \(y=-1\), where the original differential equation is undefined.

\(\frac{y^4}{4}+y=\frac{x^3}{3}+x+C\), on branches where \(y\ne-1\).

- Move the linear expression in \(y\) to the side with \(dy\). - Complete a square after integrating. - Preserve the forbidden dependent-variable value from the original equation.

1. Rearrange as \((y+2)\,dy=\cos x\,dx\). 2. Integration gives \(\frac{y^2}{2}+2y=\sin x+C_1\). 3. Completing the square gives \((y+2)^2=2\sin x+C\). 4. Valid branches must avoid \(y=-2\), where the original equation is undefined.

\((y+2)^2=2\sin x+C\), on branches where \(y\ne-2\).

- Move the squared factor involving \(y\) to the differential side. - Integrate the shifted power directly. - Return to the original denominator to identify the excluded value.

1. Separate variables: \((1+y)^2\,dy=(1+x)\,dx\). 2. Integrating gives \(\frac{(1+y)^3}{3}=x+\frac{x^2}{2}+C\). 3. The original equation is undefined at \(y=-1\), so a solution interval cannot include a point where the implicit relation produces \(y=-1\).

\(\frac{(1+y)^3}{3}=x+\frac{x^2}{2}+C\), with \(y\ne-1\) throughout any solution interval.

- Move the entire denominator to the side containing \(dy\). - Integrate each trigonometric term directly. - Return to the original equation to find where it is undefined.

1. Separate variables: \((1+\cos y)\,dy=\sin x\,dx\). 2. Integrating gives \(y+\sin y=-\cos x+C\). 3. The original denominator is zero when \(1+\cos y=0\), so \(y=(2k+1)\pi\), \(k\in\mathbb{Z}\), cannot occur in a solution interval.

\(y+\sin y=-\cos x+C\), with \(y\ne(2k+1)\pi\) for every integer \(k\) throughout a solution interval.

- Move the exponential factor involving \(y\) to the differential side. - Integrate before applying the logarithm. - The logarithm's argument determines the solution interval.

1. Separate variables: \(e^y\,dy=(x^2+1)\,dx\). 2. Integrating gives \(e^y=\frac{x^3}{3}+x+C\). 3. Therefore, \(y=\ln\left(\frac{x^3}{3}+x+C\right)\). 4. A solution is defined on any interval where \(\frac{x^3}{3}+x+C>0\). Since the cubic expression is strictly increasing, this is an interval to the right of its unique real zero.

\(y=\ln\left(\frac{x^3}{3}+x+C\right)\), on any interval where \(\frac{x^3}{3}+x+C>0\).

- Move the polynomial factor in \(y\) to the differential side. - Use the exponent's derivative in the x-integral. - Check monotonicity of the implicit left side.

1. Separate variables: \((1+y^2)\,dy=xe^{-x^2}\,dx\). 2. Integrating gives \(y+\frac{y^3}{3}=-\frac12e^{-x^2}+C\). 3. The derivative of the left side with respect to \(y\) is \(1+y^2>0\), so the implicit relation determines one real value of \(y\) for each real \(x\).

\(y+\frac{y^3}{3}=-\frac12e^{-x^2}+C\).

- Move both factors involving \(y\) to the differential side. - Expand the product into powers before integrating. - Preserve the domain required by the square root and denominator.

1. Separate variables: \(\sqrt{y}(1+y)\,dy=x\,dx\). 2. Expand the left side as \(y^{1/2}+y^{3/2}\). 3. Integrating gives \(\frac23y^{3/2}+\frac25y^{5/2}=\frac{x^2}{2}+C\). 4. The original equation requires \(y>0\).

\(\frac23y^{3/2}+\frac25y^{5/2}=\frac{x^2}{2}+C\), with \(y>0\).

- Move the entire power involving \(y\) to the differential side. - Use a shifted-power antiderivative and track its negative sign. - Check the original denominator for the excluded value.

1. Separate variables: \((1-y)^3\,dy=x\,dx\). 2. Integrating gives \(-\frac{(1-y)^4}{4}=\frac{x^2}{2}+C\). 3. Rewriting the constant gives \((1-y)^4=C_1-2x^2\). 4. The original equation is undefined at \(y=1\), so that value cannot occur in a solution interval.

\((1-y)^4=C-2x^2\), with \(y\ne1\) throughout a solution interval.

- Move the exponential factor to the differential side. - Integrate before applying a logarithm. - State the positivity condition for the logarithm's argument.

1. Separate variables: \(e^{2y}\,dy=(x^2+1)\,dx\). 2. Integrating gives \(\frac12e^{2y}=\frac{x^3}{3}+x+C\). 3. Rewriting the constant gives \(e^{2y}=\frac{2x^3}{3}+2x+C_1\). 4. Therefore, \(y=\frac12\ln\left(\frac{2x^3}{3}+2x+C_1\right)\). 5. A real solution interval requires \(\frac{2x^3}{3}+2x+C_1>0\).

\(y=\frac12\ln\left(\frac{2x^3}{3}+2x+C\right)\), on intervals where the logarithm's argument is positive.

- Check a proposed separation by multiplying back to reconstruct \(y'\). - Compare the reconstructed right-hand side term by term with \(x+y\). - Separation requires an algebraically equivalent product form, not a change in the equation.

1. Multiplying Casey's equation by \(y\) gives \(y'=y(x+1)=xy+y\). 2. The original equation is \(y'=x+y\), which is different because the \(x\)-term was incorrectly multiplied by \(y\). 3. Therefore, the proposed separated form is not algebraically equivalent to the original equation.

The rearrangement is invalid. It would imply \(y'=xy+y\), not \(y'=x+y\), so it changes the differential equation rather than separating it.

- A proportional decrease has a negative constant times the current concentration. - Divide by the concentration only while it is nonzero. - Use the starting value after integrating.

1. The decay equation is \(C'=-0.2C\). 2. For positive concentration, separate: \(\frac{dC}{C}=-0.2\,dt\). 3. Integration gives \(\ln C=-0.2t+A\), hence \(C=Be^{-0.2t}\). 4. The initial concentration gives \(B=5\).

\(C'(t)=-0.2C(t)\); \(\frac{dC}{C}=-0.2\,dt\); \(C(t)=5e^{-0.2t}\,\text{mg/L}\).

- Cooling depends on the difference from room temperature, not on the absolute temperature. - Separate using that temperature difference. - Apply the initial condition after integration.

1. The model is \(T'=-0.1(T-20)\), \(T(0)=80\). 2. Separate: \(\frac{dT}{T-20}=-0.1\,dt\). Integration gives \(\ln|T-20|=-0.1t+C\). 3. Hence \(T=20+Ae^{-0.1t}\) on the branch above room temperature. Using \(T(0)=80\) gives \(A=60\).

\(T'=-0.1(T-20),\quad T(0)=80\); \(\frac{dT}{T-20}=-0.1\,dt\); \(T(t)=20+60e^{-0.1t}\,\text{°C}\).

- Move the factor \(1+y\) to the same side as \(dy\). - Integrate both terms in \((1+y)\,dy\) rather than treating the factor as a constant. - Use the equivalent squared form to identify where a real branch can exist.

1. The differential equation is defined only when \(y\ne-1\). 2. Separate the variables: \((1+y)\,dy=e^x\,dx\). 3. Antidifferentiate: \(y+\frac{y^2}{2}=e^x+C\). 4. Equivalently, \((y+1)^2=2e^x+C\), after renaming the constant. Each branch must stay where \(2e^x+C>0\) and cannot cross \(y=-1\).

\(y+\frac{y^2}{2}=e^x+C\), equivalently \((y+1)^2=2e^x+C\). A solution branch is restricted to an interval on which \(2e^x+C>0\) and \(y\ne-1\).

- Multiply by \(1+y\) and by \(dx\), leaving the factor \(1/x\) on the \(x\)-side. - Use \(\ln|x|\) for the integral involving \(1/x\). - Record both original denominator restrictions: \(x\ne0\) and \(y\ne-1\).

1. The differential equation is defined only when \(x\ne0\) and \(y\ne-1\). 2. Separate the variables: \((1+y)\,dy=\frac{1}{x}\,dx\). 3. Antidifferentiate: \(y+\frac{y^2}{2}=\ln|x|+C\). 4. Solutions lie on intervals not containing \(x=0\) and on branches that do not cross \(y=-1\). The excluded value \(y=-1\) is not a solution.

\(y+\frac{y^2}{2}=\ln|x|+C\), on intervals where \(x\ne0\) and \(y\ne-1\)

- Divide by \(1+y^2\) and place \(x\) with \(dx\). - The \(y\)-integral is an inverse tangent. - When converting to a tangent formula, restrict each solution interval to avoid tangent poles.

1. Separate the variables: \(\frac{1}{1+y^2}\,dy=x\,dx\). 2. Antidifferentiate: \(\arctan y=\frac{x^2}{2}+C\). 3. Solve explicitly: \(y=\tan\left(\frac{x^2}{2}+C\right)\). 4. Each solution is defined on an interval where \(\frac{x^2}{2}+C\ne\frac{\pi}{2}+k\pi\) for every integer \(k\).

\(y=\tan\left(\frac{x^2}{2}+C\right)\), on intervals avoiding \(\frac{x^2}{2}+C=\frac{\pi}{2}+k\pi\), where \(k\in\mathbb Z\)

- Separate using \(dy/(1+y^2)=e^x\,dx\). - Integrate to an inverse-tangent relation before applying the tangent function. - The differential equation is defined for all real \((x,y)\), but each explicit tangent branch ends at its poles.

1. Separate the variables: \(\frac{1}{1+y^2}\,dy=e^x\,dx\). 2. Antidifferentiate: \(\arctan y=e^x+C\). 3. Solve explicitly: \(y=\tan(e^x+C)\). 4. The differential equation is defined for all real \(x\) and \(y\), but each solution is restricted to an interval where \(e^x+C\ne\frac{\pi}{2}+k\pi\) for every integer \(k\).

\(y=\tan(e^x+C)\), on intervals avoiding \(e^x+C=\frac{\pi}{2}+k\pi\), where \(k\in\mathbb Z\); the differential equation itself is defined for all real \(x\) and \(y\)

- Separate the two sums of squares into \(dy/(1+y^2)=dx/(1+x^2)\). - Integrate both sides with inverse tangents. - Since \(1+y^2\) never vanishes, there is no real equilibrium solution lost during separation; explicit tangent forms still require pole-free intervals.

1. Since \(1+y^2>0\), separate the variables without excluding any real value of \(y\): \(\frac{1}{1+y^2}\,dy=\frac{1}{1+x^2}\,dx\). 2. Antidifferentiate: \(\arctan y=\arctan x+C\). 3. Equivalently, \(y=\tan(\arctan x+C)\) on intervals where the tangent is defined. 4. There is no equilibrium solution because \(1+y^2\) is never \(0\) for real \(y\).

\(\arctan y=\arctan x+C\), equivalently \(y=\tan(\arctan x+C)\) on intervals where the tangent is defined; there is no equilibrium solution

- Test \(y=1\) in the original equation before dividing by \((y-1)^2\). - Integrate \((y-1)^{-2}\) carefully; its antiderivative has a negative reciprocal. - When solving explicitly, exclude points where the resulting denominator is zero.

1. For \(y\ne1\), separate the variables: \(\frac{1}{(y-1)^2}\,dy=x\,dx\). 2. Antidifferentiate: \(-\frac{1}{y-1}=\frac{x^2}{2}+C\). 3. Solve explicitly: \(y=1-\frac{1}{x^2/2+C}\), on intervals where \(x^2/2+C\ne0\). 4. Division excluded the equilibrium solution \(y=1\), so include it separately.

\(y=1-\frac{1}{x^2/2+C}\), on intervals where \(x^2/2+C\ne0\), together with the equilibrium solution \(y=1\)

- Check for constant solutions before dividing by an expression involving \(y\). - After separation, what is the antiderivative of a negative power of \(y\)? - The nonconstant explicit family is valid only where its denominator is nonzero.

1. For \(y\ne0\), separate the variables: \(\frac{1}{y^2}\,dy=e^x\,dx\). 2. Antidifferentiate: \(-\frac{1}{y}=e^x+C\). 3. Solve explicitly: \(y=-\frac{1}{e^x+C}\), on intervals where \(e^x+C\ne0\). 4. Division excluded the equilibrium solution \(y=0\), so include it separately.

\(y=-\frac{1}{e^x+C}\), on intervals where \(e^x+C\ne0\), together with the equilibrium solution \(y=0\)

- Separate the nonzero branches using \(y^{-2}\,dy=dx/(1+x^2)\). - Use the reciprocal-power antiderivative in \(y\) and the inverse-tangent antiderivative in \(x\). - Add \(y=0\) separately and restrict each nonconstant branch away from a zero denominator.

1. For \(y\ne0\), separate the variables: \(\frac{1}{y^2}\,dy=\frac{1}{1+x^2}\,dx\). 2. Antidifferentiate: \(-\frac{1}{y}=\arctan x+C\). 3. Solve explicitly: \(y=-\frac{1}{\arctan x+C}\), on intervals where \(\arctan x+C\ne0\). 4. Division excluded the equilibrium solution \(y=0\), so include it separately.

\(y=-\frac{1}{\arctan x+C}\), on intervals where \(\arctan x+C\ne0\), together with the equilibrium solution \(y=0\)

- Check \(y=0\) before dividing by \(y\). - For nonzero branches, integrate \(dy/y=\tan x\,dx\) using the logarithm of cosine. - Keep each solution on an interval where \(\cos x\ne0\).

1. On an interval where \(\cos x\ne0\) and for \(y\ne0\), separate the variables: \(\frac{1}{y}\,dy=\tan x\,dx\). 2. Antidifferentiate: \(\ln|y|=-\ln|\cos x|+C\). 3. Exponentiate to obtain \(y=C\sec x\) for \(C\ne0\). 4. The divided-out solution \(y=0\) is included by allowing \(C=0\). Each solution is defined on an interval that contains no point \(x=\frac{\pi}{2}+k\pi\).

\(y=C\sec x\), where \(C\in\mathbb R\), on any interval where \(\cos x\ne0\)

- Check \(y=0\) before dividing by \(y\). - For nonzero branches, integrate \(dy/y=\cot x\,dx\) using the logarithm of sine. - Keep each solution on an interval where \(\sin x\ne0\).

1. On an interval where \(\sin x\ne0\) and for \(y\ne0\), separate the variables: \(\frac{1}{y}\,dy=\cot x\,dx\). 2. Antidifferentiate: \(\ln|y|=\ln|\sin x|+C\). 3. Exponentiate to obtain \(y=C\sin x\) for \(C\ne0\). 4. The divided-out solution \(y=0\) is included by allowing \(C=0\). Each solution is defined on an interval containing no integer multiple of \(\pi\).

\(y=C\sin x\), where \(C\in\mathbb R\), on any interval where \(\sin x\ne0\)

- Before dividing by a factor, find the values that make it zero and check them in the original equation. - For other branches, separate the variables before integrating. - When solving for \(y\), state where the reciprocal denominator vanishes.

1. For \(y\ne-5\), separate and integrate to obtain \(-\frac{1}{y+5}=\frac{x^2}{2}+C\). 2. Solve explicitly: \(y=-5-\frac{1}{x^2/2+C}\), on intervals where \(x^2/2+C\ne0\). 3. Division removed the equilibrium solution \(y=-5\), so include it separately.

\(y=-5-\frac{1}{x^2/2+C}\), on intervals where \(x^2/2+C\ne0\), together with the equilibrium solution \(y=-5\)

- Check what remains on the \(x\)-side after moving the \(y\)-factor; do not confuse reciprocals. - Think about which power rule applies after the variables are separated. - Before dividing by a variable factor, check which constant solutions might disappear. Also check where the original equation is defined.

1. Rewrite or inspect the equation so that each variable is confined to one side. 2. Choice B is correct. Integration gives \(\ln|y|=-\frac{1}{x}+C\), so \(y=Ce^{-1/x}\) on intervals not containing \(x=0\).

Choice B is correct. Integration gives \(\ln|y|=-\frac{1}{x}+C\), so \(y=Ce^{-1/x}\) on intervals not containing \(x=0\).

- Apply tangent to the inverse-tangent relation to compare the two reported forms. - Remember that adding integer multiples of \(\pi\) can be absorbed into an arbitrary constant. - Equivalence is local on intervals where the tangent expression has no pole.

1. Compare the implicit and explicit relations obtained by separation. 2. They are equivalent because solving the first equation for \(y\) gives \(y=\tan(C-\cos x)\) on intervals where the tangent expression is defined.

They are equivalent because solving the first equation for \(y\) gives \(y=\tan(C-\cos x)\) on intervals where the tangent expression is defined.

- Check \(y=0\) before dividing by \(y\). - Once separated, consider which logarithmic antiderivatives you need. - Because the coefficient is undefined at \(x=2\), a solution interval must stay entirely on one side of that line.

1. For \(y\ne0\), separate and integrate to obtain \(\ln|y|=\ln|x-2|+C\). 2. Thus \(y=C(x-2)\) on any interval entirely to one side of \(x=2\). The differential equation is undefined at \(x=2\); \(C=0\) includes the equilibrium solution.

\(y=C(x-2)\), for \(C\in\mathbb R\), on an interval with \(x<2\) or \(x>2\). The differential equation is undefined at \(x=2\).

- Multiply by \(2\sqrt y\,dx\) to place the radical with \(dy\). - Use the power rule for \(y^{1/2}\) and for \(x\). - Preserve the stated condition \(y>0\) when interpreting the implicit family.

1. Multiply both sides by \(2\sqrt{y}\,dx\) to obtain \(2\sqrt{y}\,dy=x\,dx\). 2. Integrate: \(\frac{4}{3}y^{3/2}=\frac{x^2}{2}+C\). 3. The solution is valid only on intervals where \(y>0\).

\(\frac{4}{3}y^{3/2}=\frac{x^2}{2}+C\), with \(y>0\)

- Rewrite the quotient as \(e^xe^{-y}\) and move the factor involving \(y\) to the differential side. - Integrate the separated equation directly. - When taking a logarithm for an explicit form, require the resulting expression for \(e^y\) to be positive.

1. Rewrite or inspect the equation so that each variable is confined to one side. 2. Rewrite as \(e^y\,dy=e^x\,dx\). Integration gives \(e^y=e^x+C\), or \(y=\ln(e^x+C)\) where \(e^x+C>0\).

Rewrite as \(e^y\,dy=e^x\,dx\). Integration gives \(e^y=e^x+C\), or \(y=\ln(e^x+C)\) where \(e^x+C>0\).

- Rearrange the equation so that each side contains only one variable. - Look for a quantity whose derivative contains \(y e^{y^2}\). - Use the original restriction on \(y\) when describing valid real branches.

1. Rearrange the equation as \(y e^{y^2}\,dy=x\,dx\). 2. Integrating gives \(\frac{1}{2}e^{y^2}=\frac{1}{2}x^2+C_1\), so the family can be written as \(e^{y^2}=x^2+C\). 3. Because \(y\ne0\), real branches require \(y^2>0\), which is equivalent to \(x^2+C>1\). Each branch keeps either the positive or negative sign of \(y\).

\(e^{y^2}=x^2+C\), on intervals where \(x^2+C>1\), with separate positive and negative branches for \(y\).

- Rearrange so that all factors involving \(y\) are on one side. - Look for familiar derivative patterns on both sides. - After solving explicitly, account for values where the resulting trigonometric function is undefined.

1. Rearrange as \(\frac{dy}{1+y^2}=\frac{e^x}{1+e^x}\,dx\). 2. Integration gives \(\arctan y=\ln(1+e^x)+C\). 3. Thus \(y=\tan(\ln(1+e^x)+C)\). A maximal branch is any interval on which the tangent argument avoids \(\frac{\pi}{2}+k\pi\).

\(\arctan y=\ln(1+e^x)+C\), or \(y=\tan(\ln(1+e^x)+C)\), on intervals where the tangent expression is defined.

- Check any value excluded during division in the original differential equation. - Integrate \(\frac{1}{x\ln x}\) using \(\ln x\) as the inner expression. - After obtaining the parameterized family, test the parameter value that makes the nonconstant term vanish.

1. Dividing by \(1-y\) temporarily excludes \(y=1\), so that value must be checked in the original differential equation. 2. Substituting \(y=1\) gives \(y'=0\) and \(\frac{1-y}{x\ln x}=0\), so \(y=1\) is an equilibrium solution. 3. For nonconstant solutions, integration gives \(-\ln|1-y|=\ln(\ln x)+C_1\), hence \(y=1-\frac{C}{\ln x}\). 4. Choosing \(C=0\) gives \(y=1\), so the final family includes the equilibrium even though the separation step temporarily excluded it.

a) Division temporarily excludes \(y=1\), but the original equation shows that it is an equilibrium. b) \(y=1-\frac{C}{\ln x}\), for \(x>1\) c) The choice \(C=0\) gives \(y=1\).

- Rearrange so that the logarithmic expression in \(y\) appears with \(dy\). - Look for a derivative involving a logarithm of a logarithm. - Check separately the value that was excluded when dividing.

1. For \(y\ne1\), rearrange as \(\frac{dy}{y\ln y}=\frac{2x}{1+x^2}\,dx\). 2. Integration gives \(\ln|\ln y|=\ln(1+x^2)+C_1\). 3. Therefore, \(\ln y=C(1+x^2)\), where \(C\ne0\) for the divided case, and \(y=e^{C(1+x^2)}\). 4. Choosing \(C=0\) gives \(y=1\), so the equilibrium solution is included in the same family.

\(y=e^{C(1+x^2)}\), where \(C\) is any real constant. The equilibrium \(y=1\) occurs when \(C=0\).

- Move the exponential factor involving \(y\) to the side with \(dy\). - Recognize a logarithmic derivative involving \(1+x^2\). - Check the positivity condition needed by the final logarithm.

1. Rearrange as \(e^y\,dy=\frac{x}{1+x^2}\,dx\). 2. Integration gives \(e^y=\frac{1}{2}\ln(1+x^2)+C\). 3. Therefore, \(y=\ln\left(\frac{1}{2}\ln(1+x^2)+C\right)\). 4. A real solution requires \(\frac{1}{2}\ln(1+x^2)+C>0\) on its interval.

\(y=\ln\left(\frac{1}{2}\ln(1+x^2)+C\right)\), on intervals where \(\frac{1}{2}\ln(1+x^2)+C>0\).

- Move the entire dependent-variable denominator to the side with \(dy\). - After integrating, look for a useful completed square in \(y^2\). - Preserve the restriction from the original denominator.

1. Rearrange as \(y(1+y^2)\,dy=x\,dx\). 2. Integration gives \(\frac{y^2}{2}+\frac{y^4}{4}=\frac{x^2}{2}+C_1\). 3. Multiplying and completing a square gives \((1+y^2)^2=2x^2+C\). 4. Valid solution branches must avoid \(y=0\), because the original differential equation is undefined there.

\((1+y^2)^2=2x^2+C\), on branches for which \(y\ne0\).

- Move the exponential involving \(y\) to the side with \(dy\). - Look for a derivative involving \(e^{-x^2}\). - Check the sign required by the final logarithm.

1. Rearrange as \(e^{-y}\,dy=xe^{-x^2}\,dx\). 2. Integration gives \(-e^{-y}=-\frac{1}{2}e^{-x^2}+C_1\). 3. Thus \(e^{-y}=C+\frac{1}{2}e^{-x^2}\), and \(y=-\ln\left(C+\frac{1}{2}e^{-x^2}\right)\). 4. A real solution requires \(C+\frac{1}{2}e^{-x^2}>0\) on its interval.

\(y=-\ln\left(C+\frac{1}{2}e^{-x^2}\right)\), on intervals where \(C+\frac{1}{2}e^{-x^2}>0\).

- Move the linear expression in \(y\) to the side with \(dy\). - Integrate the negative power of \(x\) carefully. - Preserve both the excluded equilibrium and the singular x-value.

1. For \(y\ne2\), rearrange as \(\frac{dy}{y-2}=\frac{dx}{x^2}\). 2. Integration gives \(\ln|y-2|=-\frac{1}{x}+C_1\). 3. Therefore, \(y=2+Ce^{-1/x}\). 4. The equilibrium \(y=2\) is included by \(C=0\). Because the differential equation is undefined at \(x=0\), solution intervals lie within \((-\infty,0)\) or \((0,\infty)\).

\(y=2+Ce^{-1/x}\), including \(y=2\) when \(C=0\), on intervals contained in \((-\infty,0)\) or \((0,\infty)\).

- Move the power of \(y\) to the side with \(dy\). - Check the derivative of a linear expression times \(e^{-x}\). - Examine the value excluded when dividing by \(y^2\).

1. For \(y\ne0\), rearrange as \(y^{-2}\,dy=(1+x)e^{-x}\,dx\). 2. Integration gives \(-\frac{1}{y}=-(x+2)e^{-x}+C_1\). 3. Therefore, \(y=\frac{1}{C+(x+2)e^{-x}}\), on intervals where \(C+(x+2)e^{-x}\ne0\). 4. The equilibrium solution \(y=0\), excluded by division, must be listed separately.

\(y=\frac{1}{C+(x+2)e^{-x}}\), on intervals where \(C+(x+2)e^{-x}\ne0\), together with the equilibrium solution \(y=0\).

- Move the dependent variable to the side with \(dy\). - Integrate the logarithm using a familiar product-based technique. - Preserve both the logarithm's domain and the original denominator restriction.

1. Rearrange as \(y\,dy=\ln x\,dx\). 2. Integration gives \(\frac{y^2}{2}=x\ln x-x+C_1\). 3. Thus \(y^2=2x\ln x-2x+C\). 4. Valid branches require \(x>0\) and \(y\ne0\), as in the original equation.

\(y^2=2x\ln x-2x+C\), on branches where \(x>0\) and \(y\ne0\).

- Move every factor involving \(y\) to the same side as \(dy\). - One integral requires integration by parts. - Check the logarithm and the original denominator for restrictions.

1. Separate variables: \(y(1+\ln y)\,dy=(1+x^2)\,dx\). 2. Integrating gives \(\frac{y^2}{4}+\frac{y^2}{2}\ln y=x+\frac{x^3}{3}+C\). 3. The original equation requires \(y>0\) and \(1+\ln y\ne0\), so \(y\ne e^{-1}\).

\(\frac{y^2}{4}+\frac{y^2}{2}\ln y=x+\frac{x^3}{3}+C\), with \(y>0\) and \(y\ne e^{-1}\).

- Combine Diego’s two fractions over the common denominator \(1-y^2\). - Determine the coefficients in \(\frac{A}{1-y}+\frac{B}{1+y}\). - Check the values excluded when the equation is divided by \(1-y^2\).

1. The proposed partial-fraction identity is twice the correct value. The correct decomposition is \(\frac{1}{1-y^2}=\frac{1}{2}\left(\frac{1}{1-y}+\frac{1}{1+y}\right)\). 2. Separating and integrating therefore gives \(\frac{1}{2}\ln\left|\frac{1+y}{1-y}\right|=\arctan x+C\). 3. Division by \(1-y^2\) excludes \(y=1\) and \(y=-1\). 4. Substitution into the original equation shows that both excluded constants are equilibrium solutions.

a) A factor of \(\frac{1}{2}\) is missing in the partial-fraction decomposition. b) \(\frac{1}{2}\ln\left|\frac{1+y}{1-y}\right|=\arctan x+C\) c) \(y=1\) and \(y=-1\)

- Check for a constant solution before dividing by a power of \(y\). - Rewrite the dependent-variable factor with a negative exponent. - The sign inside the square root determines each branch's interval.

1. The equilibrium solution \(y=0\) satisfies the differential equation. 2. For \(y\ne0\), separate variables: \(y^{-3}\,dy=\frac{dx}{1+x^2}\). 3. Integration gives \(-\frac{1}{2y^2}=\arctan x+C_1\), or \(\frac{1}{y^2}=C-2\arctan x\). 4. Therefore, the nonconstant branches are \(y=\pm\frac{1}{\sqrt{C-2\arctan x}}\). 5. Each branch is defined on an interval where \(C-2\arctan x>0\).

\(y=0\), and \(y=\pm\frac{1}{\sqrt{C-2\arctan x}}\) on intervals where \(C-2\arctan x>0\).

- Move the entire polynomial in \(y\) to the differential side. - After integrating, study the derivative of the left-side function with respect to \(y\). - Strict monotonicity gives uniqueness without solving a cubic explicitly.

1. Separate variables: \((y^2+4y+5)\,dy=(x+1)\,dx\). 2. Integrating gives \(\frac{y^3}{3}+2y^2+5y=\frac{x^2}{2}+x+C\). 3. The derivative of the left side with respect to \(y\) is \(y^2+4y+5=(y+2)^2+1>0\). 4. Therefore, the left side is strictly increasing from \(-\infty\) to \(\infty\), so it has exactly one real \(y\) for every real right-side value.

\(\frac{y^3}{3}+2y^2+5y=\frac{x^2}{2}+x+C\). For each fixed \(C\), this relation determines exactly one real \(y\) for every real \(x\).

- Restrict attention to a branch that stays strictly between the two endpoint levels before dividing by the radical. - Recognize the antiderivative associated with \(\frac{1}{\sqrt{1-y^2}}\). - Check the endpoint values excluded by the separation step in the original equation.

1. On a branch with \(-1<y<1\), separate variables: \(\frac{dy}{\sqrt{1-y^2}}=x\,dx\). 2. Integrating gives \(\arcsin y=\frac{x^2}{2}+C\) on any interval where the branch remains strictly inside \((-1,1)\). 3. Dividing by \(\sqrt{1-y^2}\) excluded \(y=1\) and \(y=-1\). Both are constant solutions of the original equation. 4. Because the right side is not locally Lipschitz in \(y\) at \(y=\pm1\), the interior family plus the two constant solutions should not be described as exhausting every possible endpoint-joining solution.

Interior nonconstant branches satisfy \(\arcsin y=\frac{x^2}{2}+C\) on intervals where \(-1<y<1\). The constant solutions are \(y=1\) and \(y=-1\).

- Separate the quadratic expression in \(y\). - The x-integral is a logarithm of a logarithm. - Account for both the original x-domain and the poles of the tangent function.

1. Separate variables: \(\frac{dy}{1+y^2}=\frac{dx}{x\ln x}\). 2. Integrating gives \(\arctan y=\ln(\ln x)+C\). 3. Therefore, \(y=\tan(\ln(\ln x)+C)\). 4. A solution interval must lie in \(x>1\) and cannot contain a point where \(\ln(\ln x)+C=\frac{\pi}{2}+k\pi\).

\(y=\tan(\ln(\ln x)+C)\), on intervals in \(x>1\) that avoid \(\ln(\ln x)+C=\frac{\pi}{2}+k\pi\), \(k\in\mathbb{Z}\).

- Separate the shifted dependent variable. - Use the standard inverse-tangent integral for the x-side. - Check whether the equilibrium lost during division can be included in the explicit family.

1. For \(y\ne1\), separate variables: \(\frac{dy}{1-y}=\frac{dx}{x^2+4}\). 2. Integrating gives \(-\ln|1-y|=\frac12\arctan\left(\frac{x}{2}\right)+C\). 3. Solving for \(y\) gives \(y=1-C_1e^{-\frac12\arctan(x/2)}\). 4. The equilibrium \(y=1\) is included by taking \(C_1=0\). Every member is defined for all real \(x\).

\(y=1-Ce^{-\frac12\arctan(x/2)}\), on \((-\infty,\infty)\), including \(y=1\) when \(C=0\).

- Move the sine factor to the differential side. - Recall the antiderivative of the cosecant function. - Check every value excluded when dividing by \(\sin y\).

1. For \(\sin y\ne0\), separate variables: \(\csc y\,dy=(1+x^2)\,dx\). 2. Integrating gives \(\ln\left|\tan\frac{y}{2}\right|=x+\frac{x^3}{3}+C\). 3. Equivalently, nonconstant solutions satisfy \(\tan\frac{y}{2}=Ce^{x+x^3/3}\) with \(C\ne0\). 4. Division by \(\sin y\) excluded every equilibrium \(y=k\pi\), \(k\in\mathbb{Z}\).

Nonconstant solutions satisfy \(\tan\frac{y}{2}=Ce^{x+x^3/3}\), \(C\ne0\). The equilibrium solutions are \(y=k\pi\), \(k\in\mathbb{Z}\).

- Move the entire expression involving \(y\) to the differential side. - Study the derivative of the left side as a function of \(y\). - Combine monotonicity with end behavior to justify existence and uniqueness.

1. Separate variables: \((1+e^y)\,dy=(1+x^2)\,dx\). 2. Integration gives \(y+e^y=x+\frac{x^3}{3}+C\). 3. The function \(F(y)=y+e^y\) has derivative \(F'(y)=1+e^y>0\), so it is strictly increasing. 4. Also, \(F(y)\to-\infty\) as \(y\to-\infty\) and \(F(y)\to\infty\) as \(y\to\infty\). Therefore, the relation determines exactly one real \(y\) for every real right side.

\(y+e^y=x+\frac{x^3}{3}+C\). The left side is continuous, strictly increasing, and has range \((-\infty,\infty)\), so it determines one real \(y\) for every real \(x\) and \(C\).

- Move the squared cosine to the differential side using its reciprocal. - Recognize the derivative of the tangent function. - Check the values excluded by division.

1. Where \(\cos y\ne0\), separate variables: \(\sec^2y\,dy=x\,dx\). 2. Integrating gives \(\tan y=\frac{x^2}{2}+C\). 3. Thus nonconstant branches can be written as \(y=\arctan(\frac{x^2}{2}+C)+k\pi\). 4. Division by \(\cos^2y\) excluded the equilibrium solutions \(y=\frac{\pi}{2}+k\pi\), \(k\in\mathbb{Z}\).

Nonconstant solutions satisfy \(\tan y=\frac{x^2}{2}+C\). The equilibrium solutions are \(y=\frac{\pi}{2}+k\pi\), \(k\in\mathbb{Z}\).

- Separate only while the depth is positive. - Which power rule lets you integrate a reciprocal square root? - Keep track of the sign of a square root when choosing the time interval.

1. For \(h>0\), \(\frac{dh}{\sqrt h}=-0.4\,dt\). 2. Integration gives \(2\sqrt h=-0.4t+C\). With \(h(0)=9\), \(C=6\). 3. Thus \(\sqrt h=3-0.2t\), so \(h=(3-0.2t)^2\) for \(0\le t\le15\). 4. At \(t=15\), the depth reaches zero. Beyond that, \(3-0.2t<0\), which cannot equal \(\sqrt h\); the basin remains empty in the physical model.

After integration, \(2\sqrt h=6-0.4t\). Thus \(h(t)=(3-0.2t)^2\) for \(0\le t\le15\). The basin empties after \(15\,\text{h}\); continuing the squared formula would violate \(\sqrt h=3-0.2t\).

- Which factors in the rate depend on \(P\)? How can you put those on the same side as \(dP\)? - Can the rational expression be rewritten as simpler terms that you know how to integrate? - Use the starting amount to choose one solution from the family. - Compare the amount to half of the community for the final equation.

1. Separate: \(\frac{dP}{P(1-P/40)}=0.1\,dt\). 2. Use \(\frac{1}{P(1-P/40)}=\frac{1}{P}+\frac{1}{40-P}\). Integration gives \(\ln\left|\frac{P}{40-P}\right|=0.1t+C\). 3. Using \(P(0)=10\) gives \(\frac{P}{40-P}=\frac13e^{0.1t}\), so \(P(t)=\frac{40}{1+3e^{-0.1t}}\). 4. Setting \(P=20\) gives \(3e^{-0.1t}=1\), hence \(t=10\ln3\) hours.

\(\frac{dP}{P(1-P/40)}=0.1\,dt\) and \(\frac{1}{P(1-P/40)}=\frac{1}{P}+\frac{1}{40-P}\). Thus \(P(t)=\frac{40}{1+3e^{-0.1t}}\), reaching \(20\) at \(t=10\ln3\approx10.99\) hours.