Free math worksheets

- Substitute the initial x-value into the given family. - Use the initial y-value to determine the constant.

1. Substitute the initial condition into the family: \(5=Ce^0=C\). 2. Therefore, the particular solution is \(y=5e^{2x}\).

\(y=5e^{2x}\)

- Substitute the coordinates of the given point into the implicit family. - The value you obtain fixes the constant for the entire solution curve.

1. Substitute \(x=3\) and \(y=4\): \(C=3^2+4^2=25\). 2. The particular solution curve is \(x^2+y^2=25\).

\(x^2+y^2=25\)

- Separate the nonzero branches using \(dy/y=2x\,dx\). - Find the exponential family first, then apply \(y(0)=3\) to determine its multiplicative constant. - Check that the selected solution satisfies both the differential equation and the initial value.

1. Separate and antidifferentiate to obtain \(\ln|y|=x^2+C\). 2. Apply \(y(0)=3\), giving \(C=\ln3\). 3. The particular solution is \(y=3e^{x^2}\).

\(y=3e^{x^2}\)

- Separate \(dy/y=3x^2\,dx\) and integrate before using the point. - Substitute \(x=1\) and \(y=2\) into the general exponential family to determine the constant. - Use the unaltered exponent when writing the final particular solution.

1. Separate and antidifferentiate to obtain \(\ln|y|=x^3+C\). 2. Apply \(y(1)=2\), giving \(C=\ln2-1\). 3. The particular solution is \(y=2e^{x^3-1}\).

\(y=2e^{x^3-1}\)

- Integrate \(dy/y=\cos x\,dx\) to obtain a family involving \(e^{\sin x}\). - Apply the initial condition at \(x=0\), where \(\sin0=0\). - Verify that the chosen coefficient reproduces the stated initial value.

1. Separate and antidifferentiate to obtain \(\ln|y|=\sin x+C\). 2. Apply \(y(0)=4\), giving \(C=\ln4\). 3. The particular solution is \(y=4e^{\sin x}\).

\(y=4e^{\sin x}\)

- Separate and use the inverse-tangent antiderivative of \(1/(1+x^2)\). - Apply \(y(0)=5\) after exponentiating the logarithmic relation. - Note that the denominator \(1+x^2\) causes no real break in the solution interval.

1. Separate and antidifferentiate to obtain \(\ln|y|=\arctan x+C\). 2. Apply \(y(0)=5\), giving \(C=\ln5\). 3. The particular solution is \(y=5e^{\arctan x}\).

\(y=5e^{\arctan x}\)

- Integrate \(2x+1\) term by term after separating \(dy/y\). - Use \(y(0)=2\) to determine the multiplicative constant in the exponential family. - Keep the full quadratic-plus-linear exponent in the final model.

1. Separate and antidifferentiate to obtain \(\ln|y|=x^2+x+C\). 2. Apply \(y(0)=2\), giving \(C=\ln2\). 3. The particular solution is \(y=2e^{x^2+x}\).

\(y=2e^{x^2+x}\)

- Record that the differential equation is undefined at \(x=0\). - Separation gives a logarithmic relation whose nonzero solutions are linear on each side of \(x=0\). - Use \((2,6)\) to select the coefficient and the maximal interval \(x>0\).

1. Separate and antidifferentiate to obtain \(y=Cx\). 2. Apply \(y(2)=6\), giving \(C=3\). 3. The particular solution is \(y=3x\). 4. The initial point selects the stated branch, with \(x>0\).

\(y=3x\), with \(x>0\)

- Check the shifted factor \(y-1\) before separating. - Integrate \(dy/(y-1)=x\,dx\) and exponentiate. - Use \(y(0)=4\) to determine the coefficient of the exponential term.

1. Separate and antidifferentiate to obtain \(y=1+Ce^{x^2/2}\). 2. Apply \(y(0)=4\), giving \(C=3\). 3. The particular solution is \(y=1+3e^{x^2/2}\).

\(y=1+3e^{x^2/2}\)

- Multiply by \(y^2\,dx\) and integrate both power expressions. - Use \(y(0)=2\) to determine the cubic constant. - A real cube root gives a single real branch, but still check that it never reaches the excluded value \(y=0\).

1. Separate and antidifferentiate to obtain \(y^3=\frac{3}{2}x^2+C\). 2. Apply \(y(0)=2\), giving \(C=8\). 3. The particular solution is \(y=\sqrt[3]{\frac{3}{2}x^2+8}\).

\(y=\sqrt[3]{\frac{3}{2}x^2+8}\)

- Move \(3y^2\) to the differential side and integrate. - Use \(y(0)=1\) to determine the constant in the cubic relation. - Take the real cube root and check the original denominator condition.

1. Separate and antidifferentiate to obtain \(y^3=x^2+C\). 2. Apply \(y(0)=1\), giving \(C=1\). 3. The particular solution is \(y=\sqrt[3]{x^2+1}\).

\(y=\sqrt[3]{x^2+1}\)

- Separate \(dy/(y+1)=dx/(x+2)\). - Exponentiating the logarithmic relation gives a linear relation between \(y+1\) and \(x+2\) on one side of \(x=-2\). - Use \((0,3)\) and choose the maximal interval containing \(0\).

1. Separate and antidifferentiate to obtain \(y+1=C(x+2)\). 2. Apply \(y(0)=3\), giving \(C=2\). 3. The particular solution is \(y=2x+3\). 4. The initial point selects the stated branch, with \(x>-2\).

\(y=2x+3\), with \(x>-2\)

- Use the initial point as the lower limit: add the accumulated change from \(2\) to \(x\) to the value \(5\). - Integrate the polynomial integrand term by term. - Substitute \(x=2\) into the simplified expression to verify the initial condition.

1. Use the initial value as the starting value and accumulate the derivative from \(2\) to \(x\). 2. \(y=5+\int_2^x(t^2+1)\,dt=5+\frac{x^3-8}{3}+x-2=\frac{x^3}{3}+x+\frac{1}{3}\).

\(y=5+\int_2^x(t^2+1)\,dt=\frac{x^3}{3}+x+\frac{1}{3}\)

- Start with the value at \(x=\pi\) and add \(\int_{\pi}^{x}\cos t\,dt\). - Evaluate the sine antiderivative at both limits. - Check that the integral vanishes when \(x=\pi\).

1. Use the initial value as the starting value and accumulate the derivative from \(\pi\) to \(x\). 2. \(y=2+\int_\pi^x\cos t\,dt=2+\sin x\).

\(y=2+\int_\pi^x\cos t\,dt=2+\sin x\)

- Write the solution as the initial value plus the accumulation from \(0\) to \(x\). - Recognize \(1/(1+t^2)\) as the derivative of an inverse tangent. - Verify both the derivative and the value at \(x=0\).

1. Use the initial value as the starting value and accumulate the derivative from \(0\) to \(x\). 2. \(y=-1+\int_0^x\frac{1}{1+t^2}\,dt=\arctan x-1\).

\(y=-1+\int_0^x\frac{1}{1+t^2}\,dt=\arctan x-1\)

- At \(x=0\), the relation \(y+1=Ce^{3x^2}\) becomes \(y(0)+1=C\). - Use the shifted initial value, not \(y(0)\) alone, to determine \(C\). - Substitute the corrected constant back into the solution and check the initial condition.

1. Separate and integrate: \(\frac{dy}{y+1}=6x\,dx\), so \(\ln|y+1|=3x^2+C\) and \(y+1=Ce^{3x^2}\). 2. Apply \(y(0)=2\): \(3=C\). 3. Therefore, \(y=3e^{3x^2}-1\).

At \(x=0\), \(y+1=3\), so \(C=3\). The solution is \(y=3e^{3x^2}-1\).

- Integration gives a square relation, so solving for \(y\) creates two branches. - Evaluate both branches at \(x=0\). - Choose the branch whose value matches \(-2\) and keep that sign continuously on the solution interval.

1. Separate and integrate: \(y\,dy=x\,dx\), so \(y^2=x^2+C\). 2. Apply \(y(0)=-2\), giving \(C=4\). 3. Thus \(y=\pm\sqrt{x^2+4}\), and the initial value selects \(y=-\sqrt{x^2+4}\).

Integration gives \(y^2=x^2+4\). Since \(y(0)=-2\), the particular solution is \(y=-\sqrt{x^2+4}\).

- Check whether the right side \(1/y\) is defined at the initial point before separating or integrating. - A differentiable solution must satisfy the differential equation at every point of its interval, including the initial point. - Do not use a formal integrated curve to bypass an undefined initial slope.

1. The right side \(1/y\) is undefined when \(y=0\). 2. Because the proposed initial point lies outside the differential equation's domain, no differentiable solution can satisfy the initial condition.

No solution exists because the differential equation is undefined when \(y=0\), including at the proposed initial point.

- Use the accumulation formula with lower limit \(a\) and starting value \(b\). - Integrate \(2t\) from \(a\) to \(x\). - Check that substituting \(x=a\) returns \(b\).

1. Use the accumulation form \(y=b+\int_a^x2t\,dt\). 2. Evaluate the integral: \(y=b+x^2-a^2\).

Using an initial-value integral, \(y=b+\int_a^x2t\,dt=b+x^2-a^2\).

- Write velocity as \(10+\int_{3}^{t}(4-0.5u)\,du\). - Integrate the constant and linear terms before evaluating the limits. - Check both \(v'(t)\) and \(v(3)\) in the simplified formula.

1. Use the initial value as the starting velocity and accumulate acceleration from \(3\) to \(t\). 2. \(v(t)=10+\int_3^t(4-0.5u)\,du=10+4(t-3)-\frac{t^2-9}{4}=-\frac{1}{4}t^2+4t+\frac{1}{4}\).

\(v(t)=10+\int_3^t(4-0.5u)\,du=-\frac{1}{4}t^2+4t+\frac{1}{4}\)

- The lower limit must be the initial \(x\)-value so the integral is zero at \(x=1\). - Differentiate each candidate using the Fundamental Theorem of Calculus. - Select the expression that satisfies both \(y'=f(x)\) and \(y(1)=7\).

1. Use the initial-value accumulation form \(y=7+\int_1^x f(t)\,dt\). 2. By the Fundamental Theorem of Calculus, its derivative is \(f(x)\), and at \(x=1\) the integral is \(0\), so \(y(1)=7\). Therefore, choice B is correct.

Choice B. The derivative of \(7+\int_1^x f(t)\,dt\) is \(f(x)\), and at \(x=1\) the integral is \(0\), so \(y(1)=7\).

- Rearrange the equation so the expression involving \(y\) can be integrated on one side. - Use the initial value only after obtaining the one-parameter family. - Substitute the requested x-value into the particular solution.

1. Rearrange as \(e^y\,dy=(2x+1)\,dx\). 2. Integrating gives \(e^y=x^2+x+C\). 3. The condition \(y(0)=\ln 2\) gives \(C=2\), so \(y=\ln(x^2+x+2)\). 4. Substituting \(x=-1\) gives \(y(-1)=\ln 2\).

\(y(x)=\ln(x^2+x+2)\), and \(y(-1)=\ln 2\).

- Move the power of \(y\) to the differential side containing \(dy\). - Use the initial value to determine the constant and the correct fourth-root branch. - Evaluate the resulting particular solution at \(x=3\).

1. Rearrange as \(y^3\,dy=x\,dx\). 2. Integration gives \(\frac{y^4}{4}=\frac{x^2}{2}+C\), or \(y^4=2x^2+C_1\). 3. The initial condition gives \(C_1=16\). Since \(y(0)>0\), select \(y=(2x^2+16)^{1/4}\). 4. Therefore, \(y(3)=34^{1/4}\).

\(y(x)=(2x^2+16)^{1/4}\), and \(y(3)=34^{1/4}\).

- Separate the factor of \(y\) from the rational x-expression. - Recognize the derivative of \(1+x^2\) in the x-integral. - Use the initial condition before evaluating at the requested x-values.

1. Separate variables: \(\frac{dy}{y}=\frac{2x}{1+x^2}\,dx\). 2. Integrating gives \(\ln|y|=\ln(1+x^2)+C\), so \(y=C_1(1+x^2)\). 3. The initial condition gives \(5=2C_1\), so \(C_1=\frac52\). 4. Thus \(y=\frac52(1+x^2)\), so \(y(0)=\frac52\) and \(y(2)=\frac{25}{2}\).

\(y=\frac52(1+x^2)\) \(y(0)=\frac52\) \(y(2)=\frac{25}{2}\)

- Separate the shifted dependent variable. - Use the initial condition to determine the sign after exponentiating. - Evaluate the sine function at the requested inputs.

1. Separate variables: \(\frac{dy}{y+1}=\cos x\,dx\). 2. Integrating gives \(\ln|y+1|=\sin x+C\). 3. The initial condition gives \(y+1=e^{\sin x}\), so \(y=e^{\sin x}-1\). 4. The requested values are \(y\left(\frac{\pi}{2}\right)=e-1\) and \(y(\pi)=0\).

\(y=e^{\sin x}-1\) \(y\left(\frac{\pi}{2}\right)=e-1\) \(y(\pi)=0\)

- Separate the matching powers of \(x\) and \(y\). - Use the initial condition before taking a cube root. - Return to the original denominator to locate the endpoint of the maximal interval.

1. Separate variables: \(y^2\,dy=x^2\,dx\). 2. Integrating gives \(\frac{y^3}{3}=\frac{x^3}{3}+C\). 3. The initial condition gives \(y^3=x^3+1\), so \(y=(x^3+1)^{1/3}\). 4. The original equation is undefined at \(y=0\), which occurs at \(x=-1\). The maximal interval containing \(0\) is \((-1,\infty)\). 5. \(y(1)=\sqrt[3]{2}\).

\(y=(x^3+1)^{1/3}\), on \((-1,\infty)\), and \(y(1)=\sqrt[3]{2}\).

- Check the domain of the differential equation, not only the domain of the explicit formula. - Choose the connected interval containing the initial x-value. - Evaluate the one-sided limit from the explicit solution after fixing the interval.

1. The explicit formula alone does not determine the maximal interval; the original differential equation must also be defined. 2. The denominator \(1+x\) is zero at \(x=-1\), so no solution interval can cross that point. 3. The interval containing the initial x-value \(0\) is \((-1,\infty)\). 4. Using \(y=3x+2\), \(\lim_{x\to-1^+}y(x)=-1\). The finite limit does not remove the singularity in the differential equation.

a) The original differential equation is undefined at \(x=-1\), even though \(3x+2\) is defined there. b) \((-1,\infty)\) c) \(\lim_{x\to-1^+}y(x)=-1\)

- Move the exponential in \(y\) to the differential side. - Apply the initial condition before taking a logarithm. - The logarithm's positivity condition determines both endpoints.

1. Separate variables: \(e^{-y}\,dy=x\,dx\). 2. Integrating gives \(-e^{-y}=\frac{x^2}{2}+C\). 3. The initial condition gives \(e^{-y}=1-\frac{x^2}{2}\). 4. Thus \(y=-\ln\left(1-\frac{x^2}{2}\right)\). 5. The logarithm requires \(|x|<\sqrt2\), so the maximal interval is \((-\sqrt2,\sqrt2)\). 6. \(y(1)=-\ln\left(\frac12\right)=\ln2\).

\(y=-\ln\left(1-\frac{x^2}{2}\right)\), on \((-\sqrt2,\sqrt2)\), and \(y(1)=\ln2\).

- Separate the sum in the denominator without trying to isolate \(y\) first. - Use the initial condition to determine the constant in the implicit relation. - Substitute the target y-value and choose the positive x-value.

1. Separate variables: \((1+e^y)\,dy=2x\,dx\). 2. Integration gives \(y+e^y=x^2+C\). 3. The initial condition gives \(C=1\), so \(y+e^y=x^2+1\). 4. At \(y=\ln2\), the left side is \(\ln2+2\), so \(x^2=1+\ln2\). 5. The first positive x-coordinate is \(x=\sqrt{1+\ln2}\).

\(x=\sqrt{1+\ln2}\)

- Multiply by \(y\,dx\) and integrate to obtain a square relation. - Use \(y(0)=4\) to determine the constant before choosing a square-root branch. - Select the positive branch because it contains the initial value.

1. Separate and antidifferentiate to obtain \(y^2=2x^2+C\). 2. Apply \(y(0)=4\), giving \(C=16\). 3. The particular solution is \(y=\sqrt{2x^2+16}\).

\(y=\sqrt{2x^2+16}\)

- Separate using \(y\,dy=x^2\,dx\) and integrate. - Use the negative initial value to select the negative square-root branch. - Find the nearest point where the radicand reaches zero, since the original equation is undefined there.

1. Separate and antidifferentiate to obtain \(y^2=\frac{2}{3}x^3+C\). 2. Apply \(y(0)=-3\), giving \(C=9\). 3. The initial value selects the negative branch: \(y=-\sqrt{\frac{2}{3}x^3+9}\). 4. The radicand must be positive because the original equation is undefined at \(y=0\). Thus \(x>-\sqrt[3]{\frac{27}{2}}\).

\(y=-\sqrt{\frac{2}{3}x^3+9}\), on the maximal interval \(\left(-\sqrt[3]{\frac{27}{2}},\infty\right)\)

- Multiply by \(y\,dx\) and integrate the cosine term. - Use \(y(0)=2\) to determine the constant in the square relation. - Choose the positive branch and confirm that its radicand stays positive on the intended interval.

1. Separate and antidifferentiate to obtain \(y^2=2\sin x+C\). 2. Apply \(y(0)=2\), giving \(C=4\). 3. The particular solution is \(y=\sqrt{4+2\sin x}\).

\(y=\sqrt{4+2\sin x}\)

- Move \(1+y\) to the side with \(dy\) and integrate both terms. - Apply \(x=0\), \(y=1\) to the implicit quadratic relation. - When solving the quadratic for \(y\), choose the branch containing the initial value and exclude \(y=-1\).

1. Separate and antidifferentiate to obtain \(y+\frac{y^2}{2}=e^x+C\). 2. Apply \(y(0)=1\), giving \(C=\frac12\). 3. The particular solution is \(y=-1+\sqrt{2e^x+2}\).

\(y=-1+\sqrt{2e^x+2}\)

- Multiply by \(1+y^2\) and integrate to an implicit cubic relation. - Use \(y(0)=0\) to determine the integration constant. - The left side is strictly increasing in \(y\), so the implicit relation selects one real branch.

1. Separate and antidifferentiate to obtain \(y+\frac{y^3}{3}=\frac{x^2}{2}+C\). 2. Apply \(y(0)=0\), giving \(C=0\). 3. The particular solution is \(y+\frac{y^3}{3}=\frac{x^2}{2}\).

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

- Separate using \(dy/(1+y^2)=x\,dx\). - Apply the initial condition to the inverse-tangent relation before solving for \(y\). - Locate the nearest tangent poles to determine the maximal interval containing \(x=0\).

1. Separate and antidifferentiate to obtain \(\arctan y=\frac{x^2}{2}+C\). 2. Apply \(y(0)=0\), giving \(C=0\). 3. The particular solution is \(y=\tan\left(\frac{x^2}{2}\right)\). 4. The initial point selects the stated branch, with \(|x|<\sqrt{\pi}\).

\(y=\tan\left(\frac{x^2}{2}\right)\), with \(|x|<\sqrt{\pi}\)

- Separate with \(dy/(y+2)=\sin x\,dx\). - Integrate the sine term with the correct negative cosine antiderivative. - Apply \(y(0)=1\) to determine the shifted exponential coefficient.

1. Separate and antidifferentiate to obtain \(y=-2+Ce^{-\cos x}\). 2. Apply \(y(0)=1\), giving \(C=3e\). 3. The particular solution is \(y=-2+3e^{1-\cos x}\).

\(y=-2+3e^{1-\cos x}\)

- Separate using \((y-3)^{-2}\,dy=x\,dx\). - After integration, solve the reciprocal relation for \(y\). - Use \(y(0)=2\) to determine the constant and check where the final denominator could vanish.

1. Separate and antidifferentiate to obtain \(-\frac{1}{y-3}=\frac{x^2}{2}+C\). 2. Apply \(y(0)=2\), giving \(C=1\). 3. The particular solution is \(y=3-\frac{1}{1+x^2/2}\).

\(y=3-\frac{1}{1+x^2/2}\)

- Separate using \(y^{-2}\,dy=e^x\,dx\). - Integrate to a reciprocal relation before applying \(y(0)=-1\). - Keep the negative sign selected by the initial value when solving for \(y\).

1. Separate and antidifferentiate to obtain \(-\frac{1}{y}=e^x+C\). 2. Apply \(y(0)=-1\), giving \(C=0\). 3. The particular solution is \(y=-e^{-x}\).

\(y=-e^{-x}\)

- Separate \(y^{-2}\,dy=dx/(1+x^2)\). - Use \(y(0)=1\) in the reciprocal–inverse-tangent relation. - Find where the explicit denominator first becomes zero to determine the maximal interval containing \(0\).

1. Separate and antidifferentiate to obtain \(-\frac{1}{y}=\arctan x+C\). 2. Apply \(y(0)=1\), giving \(C=-1\). 3. The particular solution is \(y=\frac{1}{1-\arctan x}\). 4. The denominator first vanishes to the right when \(\arctan x=1\), so the maximal interval containing \(0\) is \((-\infty,\tan 1)\).

\(y=\frac{1}{1-\arctan x}\), on \((-\infty,\tan 1)\)

- Separate \(dy/y=\tan x\,dx\) and integrate using \(-\ln|\cos x|\). - Use \(y(0)=2\) to determine the coefficient in the secant form. - The maximal interval containing \(0\) ends at the nearest zeros of \(\cos x\).

1. Separate and antidifferentiate to obtain \(y=C\sec x\). 2. Apply \(y(0)=2\), giving \(C=2\). 3. The particular solution is \(y=2\sec x\). 4. The initial point selects the stated branch, with \(-\frac{\pi}{2}<x<\frac{\pi}{2}\).

\(y=2\sec x\), with \(-\frac{\pi}{2}<x<\frac{\pi}{2}\)

- Separate \(dy/y=\cot x\,dx\) and integrate using \(\ln|\sin x|\). - Apply the condition at \(x=\pi/2\), where \(\sin x=1\). - Keep the solution on the maximal interval between adjacent zeros of \(\sin x\) that contains \(\pi/2\).

1. Separate and antidifferentiate to obtain \(y=C\sin x\). 2. Apply \(y\left(\frac{\pi}{2}\right)=3\), giving \(C=3\). 3. The particular solution is \(y=3\sin x\). 4. The initial point selects the stated branch, with \(0<x<\pi\).

\(y=3\sin x\), with \(0<x<\pi\)

- Multiply by \(y^2+1\) and integrate both sides. - Use the initial condition to compare the same strictly increasing function evaluated at \(y\) and at \(x\). - A strictly increasing function takes equal values only at equal inputs.

1. Separate and antidifferentiate to obtain \(\frac{y^3}{3}+y=\frac{x^3}{3}+x+C\). 2. Apply \(y(0)=0\), giving \(C=0\). 3. The function \(F(u)=\frac{u^3}{3}+u\) is strictly increasing because \(F'(u)=u^2+1>0\). 4. Therefore, \(F(y)=F(x)\) implies \(y=x\).

\(y=x\)

- Multiply by \(y-1\) and integrate to a square relation. - Use \(y(0)=3\) to determine the constant. - Choose the branch with \(y-1>0\) because it contains the initial value.

1. Separate and antidifferentiate to obtain \(\frac{(y-1)^2}{2}=x^2+C\). 2. Apply \(y(0)=3\), giving \(C=2\). 3. The particular solution is \(y=1+\sqrt{2x^2+4}\).

\(y=1+\sqrt{2x^2+4}\)

- Move \(1+y\) to the side with \(dy\) and integrate the sine term. - Use \(y(0)=0\) to determine the constant in the quadratic relation. - Choose the square-root branch that gives \(y=0\) at \(x=0\) and avoids \(y=-1\).

1. Separate and antidifferentiate to obtain \(y+\frac{y^2}{2}=1-\cos x+C\). 2. Apply \(y(0)=0\), giving \(C=0\). 3. The particular solution is \(y=-1+\sqrt{3-2\cos x}\).

\(y=-1+\sqrt{3-2\cos x}\)

- Record that the differential equation is undefined at \(x=0\). - Separate \(dy/(1+y)=dx/x^2\) and integrate the reciprocal power. - Use \((1,0)\) to determine the constant and keep the solution on the interval \(x>0\).

1. Separate and antidifferentiate to obtain \(\ln|1+y|=-\frac{1}{x}+C\). 2. Apply \(y(1)=0\), giving \(C=1\). 3. The particular solution is \(y=e^{1-1/x}-1\). 4. The initial point selects the stated branch, with \(x>0\).

\(y=e^{1-1/x}-1\), with \(x>0\)

- Separate \(dy/(y-4)=e^{-x}\,dx\). - Integrate \(e^{-x}\) with its negative sign, then exponentiate. - Use \(y(0)=5\) to determine the positive shifted coefficient.

1. Separate and antidifferentiate to obtain \(\ln|y-4|=-e^{-x}+C\). 2. Apply \(y(0)=5\), giving \(C=1\). 3. The particular solution is \(y=4+e^{1-e^{-x}}\).

\(y=4+e^{1-e^{-x}}\)

- Integrate before completing the square in \(y\). - Use the initial condition to determine both the constant and the correct square-root sign. - After finding the branch, evaluate it at the two requested x-values.

1. Separate variables: \((1+y)\,dy=2x\,dx\). 2. Integration gives \(y+\frac{y^2}{2}=x^2+C\). 3. Using \(y(1)=3\) gives \(C=\frac{13}{2}\), so \((y+1)^2=2x^2+14\). 4. The initial condition selects \(y=-1+\sqrt{2x^2+14}\). 5. Therefore, \(y(0)=\sqrt{14}-1\), and because the formula depends on \(x^2\), \(y(-1)=3\).

Implicit relation: \(y+\frac{y^2}{2}=x^2+C\) Constant: \(C=\frac{13}{2}\) Selected branch: \(y=-1+\sqrt{2x^2+14}\) \(y(0)=\sqrt{14}-1\) \(y(-1)=3\)

- Rearrange so that the exponential in \(y\) can be integrated. - Use the initial condition before taking a logarithm. - Choose the interval on which the resulting linear factor has the sign matching the initial point.

1. Rearrange as \(e^{2y}\,dy=(1+x)\,dx\). 2. Integration gives \(\frac{1}{2}e^{2y}=x+\frac{x^2}{2}+C\). 3. The initial condition gives \(e^{2y}=(x+1)^2\), so the branch containing \(x=0\) is \(y=\ln(x+1)\). 4. The maximal interval containing \(0\) is \((-1,\infty)\), and \(y(1)=\ln2\).

\(y(x)=\ln(x+1)\), with maximal interval \((-1,\infty)\), and \(y(1)=\ln2\).

- Move the linear expression in \(y\) to the side with \(dy\). - Use the initial condition before solving the resulting quadratic for \(y\). - Let the initial condition determine which quadratic branch is valid.

1. Rearrange as \((2y+3)\,dy=e^x\,dx\). 2. Integration gives \(y^2+3y=e^x+C\). 3. The initial condition gives \(4=1+C\), so \(C=3\). 4. Solving the quadratic and using \(y(0)=1\) gives \(y=\frac{-3+\sqrt{21+4e^x}}{2}\). 5. Therefore, \(y(\ln4)=\frac{-3+\sqrt{37}}{2}\).

\(y(x)=\frac{-3+\sqrt{21+4e^x}}{2}\), and \(y(\ln4)=\frac{-3+\sqrt{37}}{2}\).

- First obtain the separated one-parameter family. - Use the initial point to select the constant and the side of the singularity. - Evaluate the exponent's limit to find the long-term function value.

1. Separation gives the family \(y=2+Ce^{-1/x}\) on intervals not crossing \(x=0\). 2. The initial condition gives \(5=2+Ce^{-1}\), so \(C=3e\). 3. Thus \(y=2+3e^{1-1/x}\), with maximal interval \((0,\infty)\). 4. As \(x\to\infty\), \(1-\frac{1}{x}\to1\), so \(y\to2+3e\).

\(y(x)=2+3e^{1-1/x}\), with maximal interval \((0,\infty)\), and \(\lim_{x\to\infty}y(x)=2+3e\).

- Move the dependent variable to the side with \(dy\). - Use the initial condition to choose the correct square-root branch. - Require the original denominator to remain nonzero when finding the interval.

1. Rearrange as \(y\,dy=e^{-x}\,dx\). 2. Integration gives \(\frac{y^2}{2}=-e^{-x}+C\). 3. The initial condition gives \(2=-1+C\), so \(y^2=6-2e^{-x}\). 4. The positive initial value selects \(y=\sqrt{6-2e^{-x}}\). The original denominator requires \(6-2e^{-x}>0\), so \(x>-\ln3\). 5. Hence the maximal interval is \((-\ln3,\infty)\), and \(y(\ln2)=\sqrt{5}\).

\(y(x)=\sqrt{6-2e^{-x}}\), with maximal interval \((-\ln3,\infty)\), and \(y(\ln2)=\sqrt{5}\).

- Move the power of \(y\) to the side with \(dy\). - Use the initial condition before taking a cube root. - Check where the original denominator becomes zero.

1. Rearrange as \(3y^2\,dy=2x\,dx\). 2. Integration gives \(y^3=x^2+C\). 3. The initial condition gives \(C=0\), so on the interval containing \(1\), \(y=x^{2/3}\). 4. The original denominator is zero at \(x=0\), so the maximal interval containing \(1\) is \((0,\infty)\). 5. Therefore, \(y(8)=8^{2/3}=4\).

\(y(x)=x^{2/3}\), with maximal interval \((0,\infty)\), and \(y(8)=4\).

- Use the separated implicit family before applying the initial condition. - Select the branch matching the positive initial value. - Analyze the radicand's minimum to determine whether it can reach zero.

1. Separation gives \(y^2=2x\ln x-2x+C\). 2. The initial condition gives \(2=-2+C\), so \(C=4\). 3. Since the initial value is positive, \(y=\sqrt{2x\ln x-2x+4}\). 4. The radicand has derivative \(2\ln x\), so its minimum on \(x>0\) occurs at \(x=1\), where it equals \(2\). Thus the maximal interval is \((0,\infty)\). 5. At \(x=e\), \(y(e)=\sqrt{2e-2e+4}=2\).

\(y(x)=\sqrt{2x\ln x-2x+4}\), with maximal interval \((0,\infty)\), and \(y(e)=2\).

- Separate and integrate before applying the initial condition. - Examine whether the left side is one-to-one and covers all real values. - Use the implicit relation first, then the differential equation, at \(x=\pi\).

1. Rearrange as \((1+y^2)\,dy=\cos x\,dx\). 2. Integration and the initial condition give \(y+\frac{y^3}{3}=\sin x\). 3. The function \(F(y)=y+\frac{y^3}{3}\) is strictly increasing and has range all real numbers, so the relation defines one real \(y\) for every real \(x\). 4. At \(x=\pi\), \(F(y)=0\), so \(y(\pi)=0\). 5. The differential equation gives \(y'(\pi)=\frac{\cos\pi}{1+0^2}=-1\).

\(y+\frac{y^3}{3}=\sin x\) for all real \(x\); \(y(\pi)=0\) and \(y'(\pi)=-1\).

- Move both dependent-variable terms to the side with \(dy\). - Use the initial condition in the implicit relation. - At \(x=1\), use monotonicity of the left side to identify the unique y-value.

1. Separate variables: \((e^y+2)\,dy=(3x^2-2x)\,dx\). 2. Integration gives \(e^y+2y=x^3-x^2+C\). 3. The initial condition gives \(C=1\), so \(e^y+2y=x^3-x^2+1\). 4. At \(x=1\), the right side is \(1\). The function \(e^y+2y\) is strictly increasing because its derivative is \(e^y+2>0\), and \(y=0\) gives value \(1\). Therefore, \(y(1)=0\).

\(e^y+2y=x^3-x^2+1\), and \(y(1)=0\).

- Separate and integrate before using either given value of \(y\). - Use the initial condition to fix the constant. - Substitute the target value into the implicit relation and choose the positive x-coordinate.

1. Separate variables: \((1+y^2)\,dy=x\,dx\). 2. Integrating gives \(y+\frac{y^3}{3}=\frac{x^2}{2}+C\). 3. The initial condition gives \(C=\frac{4}{3}\). 4. Set \(y=2\): \(2+\frac{8}{3}=\frac{x^2}{2}+\frac{4}{3}\). 5. Thus \(\frac{x^2}{2}=\frac{10}{3}\), so the first positive x-coordinate is \(x=\sqrt{\frac{20}{3}}\).

\(x=\sqrt{\frac{20}{3}}=\frac{2\sqrt{15}}{3}\).

- Separate the factor containing \(y\) before integrating. - Use the sign of the initial value to select a branch. - After finding the particular solution, use the range of \(\sin x\) to determine the extreme y-values.

1. Separate variables: \(2y\,dy=\cos x\,dx\). 2. Integrating gives \(y^2=\sin x+C\). 3. The initial condition gives \(C=4\), and the positive initial value selects \(y=\sqrt{4+\sin x}\). 4. Since \(-1\le\sin x\le1\), the expression \(4+\sin x\) ranges from \(3\) to \(5\). 5. Therefore, the minimum value of \(y\) is \(\sqrt{3}\), and the maximum value is \(\sqrt{5}\).

\(y=\sqrt{4+\sin x}\) Minimum value: \(\sqrt{3}\) Maximum value: \(\sqrt{5}\)

- Both separated integrals lead to inverse tangent expressions. - Use the initial condition before applying a tangent identity. - The explicit denominator determines the interval containing the initial point.

1. Separate variables: \(\frac{dy}{1+y^2}=\frac{dx}{1+x^2}\). 2. Integrating gives \(\arctan y=\arctan x+C\). 3. The initial condition gives \(C=\frac{\pi}{4}\). 4. Using the tangent addition formula, \(y=\frac{x+1}{1-x}\). 5. The interval containing \(0\) ends at the singularity \(x=1\), so the maximal interval is \((-\infty,1)\). 6. Substitution gives \(y(-1)=0\).

\(y=\frac{x+1}{1-x}\), on \((-\infty,1)\), and \(y(-1)=0\).

- Separate and integrate to determine which candidate forms are possible. - Use the initial condition to choose between the two square-root signs. - Check whether the selected branch can reach the singular level \(y=2\).

1. Separation gives \((y-2)^2=x^2+1\), so the only possible branches are A and B. 2. Choice A gives \(y(0)=3\), while choice B gives \(y(0)=1\). Therefore, A is the IVP solution. 3. For A, \(y-2=\sqrt{x^2+1}>0\) for every real \(x\), so the differential equation remains defined on \((-\infty,\infty)\). 4. Choice B has the opposite square-root sign. It satisfies the same squared relation and differential equation, but it does not satisfy the stated initial condition. 5. \(y(\sqrt3)=2+\sqrt4=4\).

a) A. \(y=2+\sqrt{x^2+1}\) b) \((-\infty,\infty)\) c) Choice B gives \(y(0)=1\), not \(3\). d) \(y(\sqrt3)=4\)

- Separate and integrate before selecting a branch. - Find where the branch reaches the value that makes the original denominator zero. - Take the limit in the implicit relation rather than solving explicitly first.

1. Separate variables: \((1+y)\,dy=e^{-x}\,dx\). 2. Integrating gives \(y+\frac{y^2}{2}=-e^{-x}+C\). 3. The initial condition gives \(C=\frac52\). 4. The branch containing \(y=1\) reaches the singular level \(y=-1\) when the right side equals \(-\frac12\). This occurs when \(e^{-x}=3\), or \(x=-\ln3\). 5. Hence the maximal interval is \((-\ln3,\infty)\). 6. As \(x\to\infty\), the relation becomes \(y+\frac{y^2}{2}=\frac52\). The branch containing positive values gives \(y=-1+\sqrt6\).

\(y+\frac{y^2}{2}=\frac52-e^{-x}\), on \((-\ln3,\infty)\), and \(\lim_{x\to\infty}y(x)=-1+\sqrt6\).

- Separate and integrate the cosine factor. - Use the initial point to select the inverse-sine branch. - Find where the branch reaches a value that makes the original denominator zero.

1. Separate variables: \(\cos y\,dy=(x+1)\,dx\). 2. Integrating and using the initial condition gives \(\sin y=\frac{x^2}{2}+x\). 3. The branch through \((0,0)\) is \(y=\arcsin\left(\frac{x^2}{2}+x\right)\). 4. It reaches the singular level \(y=\frac{\pi}{2}\) when \(\frac{x^2}{2}+x=1\), at \(x=-1\pm\sqrt3\). 5. Therefore, the maximal interval containing \(0\) is \((-1-\sqrt3,-1+\sqrt3)\). 6. At \(x=-1\), \(\sin y=-\frac12\), and the selected branch gives \(y(-1)=-\frac{\pi}{6}\).

\(\sin y=\frac{x^2}{2}+x\), with maximal interval \((-1-\sqrt3,-1+\sqrt3)\), and \(y(-1)=-\frac{\pi}{6}\).

- Separate using the two factors of \(1-y^2\). - Exponentiate the implicit relation before solving for \(y\). - Preserve the interval determined by the initial x-value and the singular point \(x=0\).

1. Separate variables: \(\frac{dy}{1-y^2}=\frac{dx}{x}\). 2. Integrating and using the initial condition gives \(\frac12\ln\left(\frac{1+y}{1-y}\right)=\ln x\) for \(x>0\). 3. Thus \(\frac{1+y}{1-y}=x^2\), so \(y=\frac{x^2-1}{x^2+1}\). 4. The maximal interval containing \(1\) is \((0,\infty)\). 5. \(y(2)=\frac35\), and \(\lim_{x\to0^+}y(x)=-1\).

\(y=\frac{x^2-1}{x^2+1}\), on \((0,\infty)\); \(y(2)=\frac35\), and \(\lim_{x\to0^+}y(x)=-1\).

- A logarithmic substitution removes the factor of \(y\). - Solve the resulting linear separable equation for the new variable. - Use the endpoint limits of the inverse tangent function.

1. Let \(u=\ln y\). Since \(u'=\frac{y'}{y}\), the equation becomes \(u'=\frac{u}{1+x^2}\). 2. Separating gives \(\frac{du}{u}=\frac{dx}{1+x^2}\), so \(u=Ce^{\arctan x}\). 3. The initial condition \(u(0)=1\) gives \(C=1\). 4. Therefore, \(y=\exp(e^{\arctan x})\), which is positive and defined for all real \(x\). 5. The limits are \(\exp(e^{\pi/2})\) as \(x\to\infty\) and \(\exp(e^{-\pi/2})\) as \(x\to-\infty\).

\(y=\exp(e^{\arctan x})\), on \((-\infty,\infty)\). \(\lim_{x\to\infty}y=\exp(e^{\pi/2})\), and \(\lim_{x\to-\infty}y=\exp(e^{-\pi/2})\).

- Separate the linear factor involving \(y\). - Complete the square after applying the initial condition. - Use the initial value to select a branch and the original denominator to set the interval.

1. Separate variables: \((1-y)\,dy=2x\,dx\). 2. Integrating and using the initial condition gives \(y-\frac{y^2}{2}=x^2\). 3. Rearranging gives \((y-1)^2=1-2x^2\). 4. The initial value selects \(y=1-\sqrt{1-2x^2}\). 5. The branch reaches the singular level \(y=1\) at \(x=\pm\frac{1}{\sqrt2}\), so the maximal interval is \(\left(-\frac{1}{\sqrt2},\frac{1}{\sqrt2}\right)\). 6. \(y\left(\frac12\right)=1-\sqrt{\frac12}=1-\frac{1}{\sqrt2}\).

\(y=1-\sqrt{1-2x^2}\), on \(\left(-\frac{1}{\sqrt2},\frac{1}{\sqrt2}\right)\), and \(y\left(\frac12\right)=1-\frac{1}{\sqrt2}\).

- Separate using the inverse-tangent derivative. - Use the initial condition before applying the tangent function. - Locate the nearest tangent pole in each direction from the initial point.

1. Separate variables: \(\frac{dy}{1+y^2}=e^{-x}\,dx\). 2. Integrating and using the initial condition gives \(\arctan y=1-e^{-x}\). 3. Thus \(y=\tan(1-e^{-x})\). 4. Moving left from \(0\), the first tangent pole occurs when \(1-e^{-x}=-\frac{\pi}{2}\), so \(x=-\ln\left(1+\frac{\pi}{2}\right)\). 5. Moving right, the argument approaches \(1<\frac{\pi}{2}\), so there is no right endpoint. The maximal interval is \(\left(-\ln\left(1+\frac{\pi}{2}\right),\infty\right)\). 6. The limit is \(\tan1\).

\(y=\tan(1-e^{-x})\), on \(\left(-\ln\left(1+\frac{\pi}{2}\right),\infty\right)\), and \(\lim_{x\to\infty}y(x)=\tan1\).

- Use the target y-value in the implicit particular solution. - The singular levels are determined by the denominator in the original equation. - Find the nearest x-values where the implicit relation reaches those levels.

1. Separate variables: \((1-y^2)\,dy=\cos x\,dx\). 2. Integrating and using the initial condition gives \(y-\frac{y^3}{3}=\sin x\). 3. At \(y=\frac12\), the left side is \(\frac12-\frac{1}{24}=\frac{11}{24}\), so the first positive x-coordinate is \(x=\arcsin\frac{11}{24}\). 4. The original equation is singular at \(y=\pm1\). The implicit left side equals \(\pm\frac23\) there. 5. Starting from \(x=0\), the first such points occur at \(x=\pm\arcsin\frac23\), so the maximal interval is \(\left(-\arcsin\frac23,\arcsin\frac23\right)\).

\(x=\arcsin\frac{11}{24}\), and the maximal interval is \(\left(-\arcsin\frac23,\arcsin\frac23\right)\).

- Integrate \(dy/(1+y^2)=e^x\,dx\). - Use \(y(0)=0\) to set the constant in the inverse-tangent relation. - Determine the maximal interval by finding where the tangent argument first reaches a pole on each side of the initial point.

1. Separate and antidifferentiate to obtain \(\arctan y=e^x+C\). 2. Apply \(y(0)=0\), giving \(C=-1\). 3. The particular solution is \(y=\tan(e^x-1)\). 4. The nearest tangent pole to the right satisfies \(e^x-1=\frac{\pi}{2}\); there is no finite left endpoint. The maximal interval containing \(0\) is \(\left(-\infty,\ln\left(1+\frac{\pi}{2}\right)\right)\).

\(y=\tan(e^x-1)\), on \(\left(-\infty,\ln\left(1+\frac{\pi}{2}\right)\right)\)

- Multiply by \(2y+1\) and integrate to an implicit quadratic relation. - Use \((0,0)\) before applying the quadratic formula. - Choose the quadratic branch containing \(y=0\) and find the nearest point where the original denominator \(2y+1\) becomes zero.

1. Separate and antidifferentiate to obtain \(y^2+y=\frac{x^2}{2}+x+C\). 2. Apply \(y(0)=0\), giving \(C=0\). 3. The initial condition selects \(y=\frac{-1+\sqrt{2x^2+4x+1}}{2}\). 4. The radicand is positive on the interval containing \(0\) when \(x>-1+\frac{\sqrt{2}}{2}\).

\(y=\frac{-1+\sqrt{2x^2+4x+1}}{2}\), on \(\left(-1+\frac{\sqrt{2}}{2},\infty\right)\)

- Multiply by \(2y+1\) and integrate \(3x^2\). - Use \((0,0)\) before solving the quadratic relation for \(y\). - Choose the branch containing the initial value and stop where \(2y+1=0\).

1. Separate and antidifferentiate to obtain \(y^2+y=x^3+C\). 2. Apply \(y(0)=0\), giving \(C=0\). 3. The particular solution is \(y=\frac{-1+\sqrt{1+4x^3}}{2}\). 4. The initial point selects the stated branch, with \(x>-\sqrt[3]{\frac14}\).

\(y=\frac{-1+\sqrt{1+4x^3}}{2}\), with \(x>-\sqrt[3]{\frac14}\)

- Rearrange the equation so each side contains only one variable. - Apply the initial condition to the integrated relation before solving for \(y\). - Locate where the denominator of the explicit solution vanishes and determine which side contains the initial point.

1. Rearrange as \(\frac{dy}{(1-y)^2}=(1+x^2)\,dx\). 2. Integration gives \(\frac{1}{1-y}=x+\frac{x^3}{3}+C\). 3. The initial condition gives \(-\frac{4}{3}=C\), so \(y=1-\frac{1}{x+\frac{x^3}{3}-\frac{4}{3}}\). 4. The denominator is zero at \(x=1\). Because its derivative is \(1+x^2>0\), this is its only real zero. The maximal interval containing \(0\) is \((-\infty,1)\).

\(y(x)=1-\frac{1}{x+\frac{x^3}{3}-\frac{4}{3}}\), with maximal interval \((-\infty,1)\).

- Differentiate the proposed branch and simplify \(2y-1\). - Check the original denominator at the endpoint, not only the square-root domain. - A maximal solution interval must be open and cannot contain a singular point of the differential equation.

1. Differentiating gives \(y'=\frac{3x^2}{\sqrt{4x^3+5}}\). 2. Since \(2y-1=\sqrt{4x^3+5}\), the differential equation gives the same derivative wherever the denominator is nonzero. 3. At \(x=1\), \(y(1)=\frac{1+3}{2}=2\), so the initial condition is satisfied. 4. At \(x=-\sqrt[3]{\frac{5}{4}}\), the square root is zero, so \(2y-1=0\). The original differential equation is undefined there. 5. Therefore, the maximal interval containing \(x=1\) is \(\left(-\sqrt[3]{\frac{5}{4}},\infty\right)\).

a) The formula satisfies the differential equation where \(4x^3+5>0\), and \(y(1)=2\). b) At the proposed endpoint, \(2y-1=0\), so the differential equation is undefined. c) \(\left(-\sqrt[3]{\frac{5}{4}},\infty\right)\)

- A logarithmic substitution simplifies the factor \(y'/y\). - Complete the square in the transformed variable. - Use the initial condition to select a branch and the original denominator to find the endpoint.

1. Let \(u=\ln y\). Then \(u'=\frac{y'}{y}=\frac{1}{x(1+u)}\). 2. Separate and integrate: \((1+u)\,du=\frac{dx}{x}\), so \(u+\frac{u^2}{2}=\ln x+C\). 3. The initial condition \(u(1)=1\) gives \(C=\frac32\). 4. Thus \((u+1)^2=2\ln x+4\). The initial value selects \(u=-1+\sqrt{2\ln x+4}\). 5. The solution reaches the singular level \(u=-1\) when \(x=e^{-2}\), so the maximal interval is \((e^{-2},\infty)\). 6. At \(x=e^{5/2}\), \(u=-1+\sqrt9=2\), so \(y=e^2\).

\(y=\exp\left(-1+\sqrt{2\ln x+4}\right)\), on \((e^{-2},\infty)\), and \(y(e^{5/2})=e^2\).