Free math worksheets

- Differentiate \(e^{3x}\) using the chain rule, including the factor from \(3x\). - Compute \(3y\) from the proposed function before comparing it with \(y'\). - Confirm that the equality holds for every real \(x\).

1. Differentiate the candidate: \(y'=21e^{3x}=3y\). 2. The derivative equals the right side of \(y'=3y\) wherever the candidate is defined. 3. Therefore, the candidate is a solution.

Yes; the candidate satisfies the differential equation on its domain.

- Recall that a differential equation must hold throughout an interval. - Differentiate the proposed function before comparing it with the right side. - Substitute \(x=0\) into the function, not into its derivative, for the initial value.

1. A solution must satisfy the differential equation for every \(x\) in its interval, not only at one selected point. 2. Differentiating \(y=x^2+1\) gives \(y'=2x\), which matches the right side for every real \(x\). 3. At \(x=0\), \(y(0)=0^2+1=1\), so the corresponding initial condition is \(y(0)=1\).

a) One matching point does not establish equality for all \(x\). b) Since \(y'=2x\) for every real \(x\), the function is a solution on \((-\infty,\infty)\). c) \(y(0)=1\)

- Differentiate the reciprocal quadratic carefully. - Keep a common denominator when substituting. - Compare the two terms before combining them.

1. Differentiate to obtain \(y'=-\frac{2Cx}{(1+x^2)^2}\). 2. Then \((1+x^2)y'=-\frac{2Cx}{1+x^2}\). 3. Also \(2xy=\frac{2Cx}{1+x^2}\). 4. The two terms sum to \(0\), so the family satisfies the equation for every real \(x\).

The family satisfies \((1+x^2)y'+2xy=0\) for all real \(x\).

- Differentiate both exponential layers carefully. - Replace the unchanged exponential product by \(y\). - Move all terms to one side for the final check.

1. Differentiate using the chain rule: \(y'=Ce^{-e^x}(-e^x)\). 2. Since \(Ce^{-e^x}=y\), this is \(y'=-e^xy\). 3. Therefore, \(y'+e^xy=0\) for every real \(x\).

The family satisfies \(y'+e^xy=0\) for all real \(x\).

- Apply the chain rule to the exponent \(x^2\). - Replace the unchanged exponential product by \(y\). - Compare the resulting derivative with Rina's equation.

1. Differentiate using the chain rule: \(y'=Ce^{x^2}(2x)\). 2. Since \(Ce^{x^2}=y\), this becomes \(y'=2xy\). 3. Rina omitted the factor \(2\) from differentiating \(x^2\). The family satisfies \(y'=2xy\), not \(y'=xy\).

Rina's claim is false. The family satisfies \(y'=2xy\).

- Differentiate the quotient before substituting. - Simplify the multiplied derivative separately. - Identify the denominator's excluded x-value.

1. Differentiate: \(y'=\frac{C}{(1+x)^2}\). 2. Then \(x(1+x)y'=\frac{Cx}{1+x}=y\). 3. Therefore, \(x(1+x)y'-y=0\). 4. The formula is valid on intervals that do not contain \(x=-1\): \((-\infty,-1)\) or \((-1,\infty)\).

The family satisfies the differential equation on \((-\infty,-1)\) and on \((-1,\infty)\).

- Differentiate both the polynomial and reciprocal terms. - Substitute before combining like terms. - Identify the x-value excluded by the formula.

1. Differentiate: \(y'=\frac{2x}{3}-\frac{C}{x^2}\). 2. Then \(xy'+y=\left(\frac{2x^2}{3}-\frac{C}{x}\right)+\left(\frac{x^2}{3}+\frac{C}{x}\right)=x^2\). 3. The formula is defined on intervals not containing \(x=0\).

The family satisfies the equation on \((-\infty,0)\) and on \((0,\infty)\).

- Apply the chain rule to the exponential. - Replace the repeated exponential product by \(y\). - Multiply before checking the final cancellation.

1. Differentiate: \(y'=Ce^{\arctan x}\frac{1}{1+x^2}\). 2. Since \(Ce^{\arctan x}=y\), \(y'=\frac{y}{1+x^2}\). 3. Therefore, \((1+x^2)y'-y=0\) for every real \(x\).

The family satisfies \((1+x^2)y'-y=0\) for all real \(x\).

- Use the product rule on the polynomial and logarithmic factors. - Multiply the derivative by \(x\) before subtracting. - Use the logarithm to state the domain.

1. Differentiate: \(y'=2x(C+\ln x)+x\). 2. Then \(xy'=2x^2(C+\ln x)+x^2\). 3. Subtracting \(2y=2x^2(C+\ln x)\) leaves \(x^2\).

The family satisfies \(xy'-2y=x^2\) for \(x>0\).

- Apply the product rule to the trigonometric and exponential factors. - Recognize the unchanged product as the original function. - Rearrange before checking the equation.

1. Differentiate using the product rule: \(y'=e^{-x}\cos x-(C+\sin x)e^{-x}\). 2. The second term is \(-y\), so \(y'=e^{-x}\cos x-y\). 3. Therefore, \(y'+y=e^{-x}\cos x\) for every real \(x\).

The family satisfies \(y'+y=e^{-x}\cos x\) for all real \(x\).

- Apply the chain rule to the exponent \(x^2\) when differentiating \(Ce^{x^2}\). - Rewrite the right side \(2xy\) by substituting the entire candidate for \(y\). - Compare the two expressions symbolically for arbitrary \(C\), rather than checking one value of \(x\).

1. Differentiate the candidate: \(y'=2xCe^{x^2}=2xy\). 2. The derivative equals the right side of \(y'=2xy\) wherever the candidate is defined. 3. Therefore, the candidate is a solution.

Yes; the candidate satisfies the differential equation on its domain.

- Differentiate each term in \(e^x-x-1\). - Substitute the candidate into \(y+x\) and simplify the cancellation involving \(x\). - The candidate is a solution only if the simplified right side equals the derivative for all \(x\).

1. Differentiate the candidate: \(y'=e^x-1\). 2. Substitute the candidate into the right side: \(y+x=(e^x-x-1)+x=e^x-1=y'\). 3. Therefore, the candidate is a solution on its domain.

Yes; the candidate satisfies the differential equation on its domain.

- Differentiate \(x+1+Ce^x\) term by term. - Compute \(y-x\) using the complete candidate, including the constant term \(1\). - Compare the resulting expressions for arbitrary \(C\).

1. Differentiate the candidate: \(y'=1+Ce^x=y-x\). 2. The derivative equals the right side of \(y'=y-x\) wherever the candidate is defined. 3. Therefore, the candidate is a solution.

Yes; the candidate satisfies the differential equation on its domain.

- Differentiate the square root with the chain rule. - Use \(y=\sqrt{x^2+4}\) to rewrite the denominator on the right side. - Check that the candidate never makes the quotient \(x/y\) undefined.

1. Differentiate the candidate: \(y'=\frac{x}{\sqrt{x^2+4}}=\frac{x}{y}\). 2. The derivative equals the right side of \(y'=x/y\) wherever the candidate is defined. 3. Therefore, the candidate is a solution.

Yes; the candidate satisfies the differential equation on its domain.

- Rewrite the candidate as \((x^2+3)^{-1}\) before differentiating. - Express \(y^2\) as \((x^2+3)^{-2}\). - Compare the derivative and \(-2xy^2\) on the candidate’s domain.

1. Differentiate the candidate: \(y'=-\frac{2x}{(x^2+3)^2}=-2xy^2\). 2. The derivative equals the right side of \(y'=-2xy^2\) wherever the candidate is defined. 3. Therefore, the candidate is a solution.

Yes; the candidate satisfies the differential equation on its domain.

- Differentiate the exponent \(x^3/3\) carefully. - Rewrite \(1+y\) from the candidate before multiplying by \(x^2\). - Compare the common exponential factor on both sides.

1. Differentiate the candidate: \(y'=4x^2e^{x^3/3}=x^2(1+y)\). 2. The derivative equals the right side of \(y'=x^2(1+y)\) wherever the candidate is defined. 3. Therefore, the candidate is a solution.

Yes; the candidate satisfies the differential equation on its domain.

- Differentiate the linear and exponential terms separately. - Substitute the complete candidate into \(2x-y\) and combine like terms. - Verify equality for every real \(x\), not only at a convenient point.

1. Differentiate the candidate: \(y'=2-5e^{-x}\). 2. Substitute the candidate into the right side: \(2x-y=2x-\bigl(2x-2+5e^{-x}\bigr)=2-5e^{-x}\). 3. The two expressions agree for every real \(x\), so the candidate is a solution.

Yes; \(y'=2-5e^{-x}=2x-y\) for every real \(x\).

- Differentiate twice because the equation contains \(y''\). - Track the factor introduced by differentiating \(\sin(2x)\) and \(\cos(2x)\) twice. - Check whether the second derivative equals \(-4y\).

1. Differentiate the candidate: \(y''=-12\cos(2x)+4\sin(2x)=-4y\). 2. The derivative equals the right side of \(y''+4y=0\) wherever the candidate is defined. 3. Therefore, the candidate is a solution.

Yes; the candidate satisfies the differential equation on its domain.

- Differentiate both exponential terms twice. - Remember that the second derivative of \(e^{-x}\) is positive \(e^{-x}\). - Compare \(y''\) directly with \(y\).

1. Differentiate the candidate: \(y''=e^x+e^{-x}=y\). 2. The derivative equals the right side of \(y''-y=0\) wherever the candidate is defined. 3. Therefore, the candidate is a solution.

Yes; the candidate satisfies the differential equation on its domain.

- Compute \(y'\) and \(2x+y\) as separate expressions. - Simplify the right side using the full candidate, including the linear terms. - One persistent mismatch is enough to refute the claim.

1. Differentiate the candidate and compare it with the right side: \(y'=-2+e^x,\quad 2x+y=2+e^x\). 2. The two expressions are not equal for all \(x\). 3. Therefore, the candidate is not a solution.

No; the derivative does not equal the required right side for all \(x\).

- Differentiate \(x^2+y^2=25\) implicitly, treating \(y\) as a function of \(x\). - Every derivative of a power of \(y\) must include a factor of \(y'\). - Isolate \(y'\) and restrict the verification to points with \(y\ne0\).

1. Differentiate the relation implicitly: \(2x+2yy'=0\). 2. Solve for the derivative to obtain \(y'=-\frac{x}{y}\). 3. This matches the differential equation at all points where its denominator is nonzero.

The relation verifies \(y'=-\frac{x}{y}\) on each differentiable branch where the differential equation is defined.

- Differentiate \(e^y\) with the chain rule, producing a factor of \(y'\). - Differentiate \(x^2+C\), remembering that \(C\) is constant. - After solving for \(y'\), rewrite \(1/e^y\) as \(e^{-y}\).

1. Differentiate the relation implicitly: \(e^y y'=2x\). 2. Solve for the derivative to obtain \(y'=2xe^{-y}\). 3. This matches the differential equation at every point on each differentiable branch of the relation.

The relation verifies \(y'=2xe^{-y}\) on each differentiable branch where the differential equation is defined.

- Differentiate both cubic terms implicitly. - Factor out the common coefficient before isolating \(y'\). - The quotient form is valid only where \(y\ne0\).

1. Differentiate the relation implicitly: \(3x^2+3y^2y'=0\). 2. Solve for the derivative to obtain \(y'=-\frac{x^2}{y^2}\). 3. This matches the differential equation at all points where its denominator is nonzero.

The relation verifies \(y'=-\frac{x^2}{y^2}\) on each differentiable branch where the differential equation is defined.

- Use the product rule on \(xy\). - Combine the two terms containing \(y'\) before solving for it. - Note the restriction created when dividing by \(x+1\).

1. Differentiate the relation implicitly: \(xy'+y+y'=0\). 2. Solve for the derivative to obtain \(y'=-\frac{y}{x+1}\). 3. This matches the differential equation at all points where its denominator is nonzero.

The relation verifies \(y'=-\frac{y}{x+1}\) on each differentiable branch where the differential equation is defined.

- Differentiate \(\ln|y|\) as \(y'/y\) on a branch where \(y\ne0\). - Differentiate \(\sin x+C\) term by term. - Multiply by \(y\) only after recording the nonzero-domain condition.

1. Differentiate the relation implicitly: \(\frac{y'}{y}=\cos x\). 2. Solve for the derivative to obtain \(y'=y\cos x\). 3. This matches the differential equation at every point on each differentiable branch, where \(y\ne0\) is already required by \(\ln|y|\).

The relation verifies \(y'=y\cos x\) on each differentiable branch where the differential equation is defined.

- Differentiate \(y^2-2y\) implicitly and collect the terms containing \(y'\). - Factor the coefficient \(2(y-1)\) from those terms. - The final quotient is valid only where \(y\ne1\).

1. Differentiate the relation implicitly: \((2y-2)y'=2x\). 2. Solve for the derivative to obtain \(y'=\frac{x}{y-1}\). 3. This matches the differential equation at all points where its denominator is nonzero.

The relation verifies \(y'=\frac{x}{y-1}\) on each differentiable branch where the differential equation is defined.

- Differentiate \(Ae^{mx}\), including the factor \(m\) from the exponent. - Use \(A\ne0\) to cancel the common exponential factor legitimately. - Match the remaining coefficient with the coefficient in \(y'=5y\).

1. Differentiate the candidate: \(y'=mAe^{mx}=my\). 2. Substitute into \(y'=5y\) and compare coefficients. 3. The required value is \(m=5\).

\(m=5\)

- Differentiate \(x^p\) using the power rule on \(x>0\). - Multiply the derivative by \(x\) before comparing with \(4y\). - Match the coefficients of the common power \(x^p\).

1. Differentiate the candidate: \(xy'=px^p=py\). 2. Substitute into \(xy'=4y,\ x>0\) and compare coefficients. 3. The required value is \(p=4\).

\(p=4\)

- Use the chain rule on the exponent \(ax^2\). - Replace \(e^{ax^2}\) by \(y\) after differentiating. - Match the coefficient of \(xy\) with the required coefficient.

1. Differentiate the candidate: \(y'=2axe^{ax^2}=2axy\). 2. Substitute into \(y'=6xy\) and compare coefficients. 3. The required value is \(a=3\).

\(a=3\)

- Differentiate \(\sin(bx)\) twice. - Express the second derivative as a multiple of the original function. - Use \(b>0\) after solving the resulting equation for \(b^2\).

1. Differentiate the candidate: \(y''=-b^2\sin(bx)=-b^2y\). 2. Substitute into \(y''+9y=0\) and compare coefficients. 3. The required value is \(b=3\).

\(b=3\)

- Differentiate only the exponential term; the added constant \(2\) disappears. - Observe that \(y-2\) equals the exponential part of the candidate. - Check whether the resulting identity imposes any restriction on \(r\).

1. Differentiate the candidate: \(y'=rKe^{rx}=r(y-2)\). 2. The result matches the differential equation for every real value of \(r\). 3. Therefore the family is valid for arbitrary \(r\).

\(r\) may be any real constant.

- Differentiate both sides term by term. - Remember the coefficient when differentiating \(2x^2\). - Note where the differential equation's denominator is nonzero.

1. The derivative of \(2x^2\) is \(4x\), not \(2x\). 2. Implicit differentiation gives \(2yy'=4x\), so \(y'=\frac{2x}{y}\). 3. Thus each differentiable branch satisfies the differential equation at points where \(y\ne0\).

Riley differentiated \(2x^2\) incorrectly. The correct calculation is \(2yy'=4x\), so \(y'=\frac{2x}{y}\); the family is valid on differentiable branches where \(y\ne0\).

- Differentiate the complete expression \(Ce^x-1\); the derivative of \(-1\) is zero. - Substitute the complete candidate into \(y+1\). - Compare the two expressions before deciding whether the constant shift causes a problem.

1. Sam's error is substituting an incomplete candidate and treating the constant shift as though it remained after differentiation. 2. Differentiate the full function: \(y'=Ce^x\). 3. Substitute the full function into the right side: \(y+1=(Ce^x-1)+1=Ce^x=y'\). Therefore, the family is valid.

Sam substituted an incomplete candidate. Using the full function gives \(y'=Ce^x\) and \(y+1=(Ce^x-1)+1=Ce^x\), so the family is valid.

- Differentiate \(x^3+C\) and note what happens to an arbitrary constant. - Compare the derivative with \(3x^2\) for every \(x\). - Different values of \(C\) may give distinct solution curves even when their derivatives agree.

1. Identify the derivative order and differentiate the complete candidate or relation. 2. The derivative of every constant is zero, so \(y'=3x^2\) for every real \(C\). The whole family works.

The derivative of every constant is zero, so \(y'=3x^2\) for every real \(C\). The whole family works.

- Differentiate \(\sin x+C\) and observe that \(C\) disappears. - Verification of the equation does not require different solution curves to have different derivatives. - Explain how changing \(C\) changes the vertical position of the curve.

1. Identify the derivative order and differentiate the complete candidate or relation. 2. Every real \(C\) gives a distinct vertical translate, and each has derivative \(\cos x\). There are infinitely many solutions.

Every real \(C\) gives a distinct vertical translate, and each has derivative \(\cos x\). There are infinitely many solutions.

- The differential equation contains \(y''\), so one derivative is not enough. - Differentiate \(x^3+Ax+B\) twice. - Track which parameter terms vanish after the second derivative.

1. Identify the derivative order and differentiate the complete candidate or relation. 2. The equation involves the second derivative. Differentiating again gives \(y''=6x\), so every \(A\) and \(B\) works.

The equation involves the second derivative. Differentiating again gives \(y''=6x\), so every \(A\) and \(B\) works.

- Begin by differentiating the accumulation function. - Compare the second derivative with the expression involving the first derivative. - Evaluate the integral when its upper and lower limits are equal.

1. By the Fundamental Theorem of Calculus, \(y'(x)=e^{x^2}\). 2. Differentiate again: \(y''(x)=2xe^{x^2}=2xy'(x)\). 3. At \(x=1\), the integral has equal limits, so \(y(1)=3\).

Yes. The function satisfies \(y''=2xy'\) and \(y(1)=3\).

- Check the initial condition first to eliminate any candidate that has the wrong value at \(x=0\). - For each remaining candidate, differentiate and then rewrite the result using that candidate’s expression for \(y\). - A function must make the two sides of the differential equation equal for every \(x\) in its interval, not just at one point.

1. For choice A, \(y'=3e^{x^2}(2x)=2x(3e^{x^2})=2xy\), and \(y(0)=3e^0=3\). Therefore, choice A satisfies both conditions. 2. For choice B, \(y'=6e^{2x}\), while \(2xy=2x(3e^{2x})=6xe^{2x}\). These expressions are not equal for all \(x\), so choice B is not a solution.

A. \(y=3e^{x^2}\). Its derivative is \(2xy\), and \(y(0)=3\). Choice B fails because \(6e^{2x}\ne6xe^{2x}\) for all \(x\).

- Rewrite \(1\) with denominator \(1+e^{-x}\) before subtracting \(y\). - Differentiate the reciprocal using the chain rule and track the two negative signs. - Verification requires the expressions for \(y'\) and \(y(1-y)\) to match and the initial value to be correct.

1. Subtracting from \(1\) gives \(1-y=1-\frac{1}{1+e^{-x}}=\frac{e^{-x}}{1+e^{-x}}\), so A is \(\frac{e^{-x}}{1+e^{-x}}\). 2. Differentiating \((1+e^{-x})^{-1}\) gives \(y'=\frac{e^{-x}}{(1+e^{-x})^2}\), so B is \(\frac{e^{-x}}{(1+e^{-x})^2}\). 3. Multiplying gives \(y(1-y)=\frac{1}{1+e^{-x}}\frac{e^{-x}}{1+e^{-x}}=\frac{e^{-x}}{(1+e^{-x})^2}\), so C equals B. 4. Also, \(y(0)=\frac{1}{1+1}=\frac{1}{2}\). Therefore, the function satisfies both the differential equation and the initial condition.

A: \(\frac{e^{-x}}{1+e^{-x}}\) B: \(\frac{e^{-x}}{(1+e^{-x})^2}\) C: \(\frac{e^{-x}}{(1+e^{-x})^2}\) Yes, the function satisfies the initial-value problem.

- Differentiate the logarithm as a composite function. - Rewrite \(e^{-y}\) using the proposed expression for \(y\). - Check the initial value separately after matching the derivative expressions.

1. The student omitted the derivative of the inside function. By the chain rule, \(y'=\frac{2x}{1+x^2}\). 2. Since \(y=\ln(1+x^2)\), \(e^{-y}=\frac{1}{1+x^2}\). Therefore, \(2xe^{-y}=\frac{2x}{1+x^2}=y'\). 3. Also, \(y(0)=\ln 1=0\). Thus the function satisfies both the differential equation and the initial condition for every real \(x\).

The student omitted the chain-rule factor \(2x\). The correct derivative is \(\frac{2x}{1+x^2}=2xe^{-y}\), and \(y(0)=0\), so the function satisfies the initial-value problem on \((-\infty,\infty)\).

- Differentiate every term with respect to \(x\), including the product \(xy\). - Collect all terms containing \(y'\) before solving for it. - State the condition needed to divide by the resulting coefficient.

1. Differentiate the relation implicitly: \(2x+xy'+y+2yy'=0\). 2. Group the derivative terms: \((x+2y)y'=-(2x+y)\). 3. Where \(x+2y\ne0\), \(y'=-\frac{2x+y}{x+2y}\), which matches the differential equation.

Yes. Every differentiable branch satisfies the equation at points where \(x+2y\ne0\).

- Differentiate the exponential using the chain rule. - Rewrite the derivative so the original expression for \(y\) appears as a factor. - Use the initial condition to solve for \(C\), and check which parameter value makes the function constant.

1. Differentiating gives \(y'=C(-2x)e^{-x^2}=-2x(Ce^{-x^2})=-2xy\), so every value of \(C\) gives a solution. 2. The condition \(y(1)=3\) gives \(Ce^{-1}=3\), so \(C=3e\). Thus \(y=3e^{1-x^2}\). 3. When \(C=0\), the family gives \(y=0\), which is the equilibrium solution.

a) \(y'=C(-2x)e^{-x^2}=-2xy\) b) \(y=3e^{1-x^2}\) c) \(y=0\), obtained when \(C=0\)

- Rewrite the candidate as a product with \(x^{-1}\) before differentiating. - Combine \(xy'\) with the original function and simplify. - Preserve the logarithm's domain restriction.

1. Differentiate using the quotient or product rule: \(y'=\frac{1-\ln x-C}{x^2}\). 2. Then \(xy'+y=\frac{1-\ln x-C}{x}+\frac{\ln x+C}{x}=\frac{1}{x}\). 3. The logarithm requires \(x>0\), so the verification holds on \((0,\infty)\).

Yes. Every member of the family satisfies the differential equation on \((0,\infty)\).

- Differentiate the parameter term carefully. - Add the derivative and the original function before simplifying. - Look for cancellation involving the arbitrary constant.

1. Differentiate: \(y'=2x+2-Ce^{-x}\). 2. Add the original function: \(y'+y=2x+2-Ce^{-x}+x^2+2x+2+Ce^{-x}\). 3. The exponential terms cancel, leaving \(x^2+4x+4\).

Yes. Every member of the family satisfies the differential equation on \((-\infty,\infty)\).

- Differentiate each candidate family with its constant parameter unchanged. - Compare \(xy'\) with \(2y\) as expressions, not at only one x-value. - Apply the initial value after identifying the correct family.

1. For choice A, \(y'=2Cx\), so \(xy'=2Cx^2=2y\). Thus choice A satisfies the differential equation for every constant \(C\). 2. The condition \(y(2)=12\) gives \(4C=12\), so \(C=3\). 3. Therefore, the required particular solution is \(y=3x^2\).

a) A. \(y=Cx^2\) b) \(y=3x^2\)

- Check the requirement for the argument of the outer logarithm, not only the inner one. - Differentiate the nested logarithm using the chain rule. - Rewrite \(e^{-y}\) by substituting the proposed formula for \(y\).

1. The outer logarithm requires \(\ln x>0\), so the correct domain is \(x>1\), not merely \(x>0\). 2. Differentiating gives \(y'=\frac{1}{x\ln x}\), so \(xy'=\frac{1}{\ln x}\). 3. Since \(y=\ln(\ln x)\), \(e^{-y}=e^{-\ln(\ln x)}=\frac{1}{\ln x}\). Thus \(xy'=e^{-y}\) for \(x>1\). 4. At \(x=e\), \(y(e)=\ln 1=0\).

a) The domain is \((1,\infty)\). b) \(xy'=\frac{1}{\ln x}=e^{-y}\) on that interval. c) \(y(e)=0\).

- Differentiate \(\ln|x|\) on an interval that does not cross zero. - Substitute the family and its derivative into the entire left side. - Identify the point excluded by the logarithmic formula and separate the valid intervals there.

1. For \(x\ne0\), \(y'=\ln|x|+C+1\). 2. Then \(xy'-y=x(\ln|x|+C+1)-x(\ln|x|+C)=x\). 3. The formula and derivative are valid separately on \((-\infty,0)\) and \((0,\infty)\).

The family satisfies the differential equation on \((-\infty,0)\) and on \((0,\infty)\).

- Differentiate the entire product, not only the exponential factor. - Substitute the derivative into both sides of Jordan's proposed equation. - Use the initial condition only after identifying the correct differential equation.

1. Differentiate with the product rule: \(y'=[1-(C+x)]e^{-x}\). 2. Then \(y'+y=[1-(C+x)+(C+x)]e^{-x}=e^{-x}\), not \(0\). Jordan's claim is false because the factor \(C+x\) also contributes to the derivative. 3. The family actually satisfies \(y'+y=e^{-x}\). 4. The condition \(y(0)=3\) gives \(C=3\), so the required member is \(y=(3+x)e^{-x}\).

Jordan's claim is false. The family satisfies \(y'+y=e^{-x}\), and the member with \(y(0)=3\) is \(y=(3+x)e^{-x}\).

- Differentiate each logarithmic candidate using the chain rule. - Rewrite the resulting fraction as \(1\) plus or minus a reciprocal. - Use the original formula to replace that reciprocal by \(e^{-y}\).

1. For choice A, \(y=\ln(Ce^x-1)\), so \(y'=\frac{Ce^x}{Ce^x-1}\). 2. Rewrite the derivative as \(y'=1+\frac{1}{Ce^x-1}\). Since \(e^y=Ce^x-1\), this is \(y'=1+e^{-y}\). 3. Thus choice A is correct. 4. The logarithm requires \(Ce^x-1>0\), and the solution is valid on any interval where that condition holds.

Choice A: \(y=\ln(Ce^x-1)\), on intervals where \(Ce^x>1\).

- Differentiate the square-root formula before substituting. - Check both the radical and the denominator in the original equation. - Use the initial condition to determine \(C\).

1. Where \(C+x^2>0\), differentiation gives \(y'=\frac{x}{\sqrt{C+x^2}}=\frac{x}{y}\), so the formula satisfies the differential equation there. 2. The original differential equation is undefined when \(y=0\). Therefore, points where \(C+x^2=0\) cannot belong to a solution interval. Noah's all-real claim is false for constants that allow such points or make the radical nonreal. 3. The initial condition gives \(2=\sqrt C\), so \(C=4\). 4. The required solution is \(y=\sqrt{x^2+4}\), which is positive and valid for every real \(x\).

The family works only on intervals where \(C+x^2>0\). For \(y(0)=2\), \(C=4\) and \(y=\sqrt{x^2+4}\) on \((-\infty,\infty)\).

- Use both the product rule and chain rule on the first term. - Collect the two terms containing \(y'\). - State the condition required before dividing.

1. Differentiate implicitly: \(e^y+xe^y y'+y'=0\). 2. Collecting derivative terms gives \((xe^y+1)y'=-e^y\). 3. Where \(xe^y+1\ne0\), division gives \(y'=-\frac{e^y}{xe^y+1}\).

Each differentiable branch satisfies the differential equation wherever \(xe^y+1\ne0\).

- Differentiate each candidate before combining \(y'\) and \(y\). - Look for cancellation of the parameter-dependent exponential terms. - Check that the remaining expression is exactly \(x\).

1. For choice B, \(y=x-1+Ce^{-x}\), so \(y'=1-Ce^{-x}\). 2. Then \(y'+y=1-Ce^{-x}+x-1+Ce^{-x}=x\). 3. Therefore, choice B satisfies the differential equation for every real \(x\). 4. Choice A gives \(y'+y=x+2\), and choice C leaves an uncanceled \(2Ce^x\) term, so neither is correct.

Choice B: \(y=x-1+Ce^{-x}\).

- Differentiate both the linear and reciprocal terms. - Substitute before deciding whether the parameter terms cancel. - Check the denominator to identify the excluded x-value.

1. Differentiate: \(y'=C-1/x^2\). 2. Then \(xy'-y=Cx-1/x-(Cx+1/x)=-2/x\). 3. Alex's claim is false. The family satisfies \(xy'-y=-2/x\). 4. The formula and differential equation are valid on intervals that do not contain \(x=0\): \((-\infty,0)\) or \((0,\infty)\).

The family satisfies \(xy'-y=-2/x\), not \(xy'-y=0\), on \((-\infty,0)\) and \((0,\infty)\).

- Apply implicit differentiation to the exponential of a product. - Collect every term containing \(y'\) before dividing. - Preserve the stated denominator condition.

1. Differentiate implicitly to obtain \(2x+2yy'+e^{xy}(y+xy')=0\). 2. Group the terms containing \(y'\): \((2y+xe^{xy})y'=-(2x+ye^{xy})\). 3. Where \(2y+xe^{xy}\ne0\), division gives \(y'=-\frac{2x+ye^{xy}}{2y+xe^{xy}}\), as required.

Each differentiable branch satisfies \(y'=-\frac{2x+ye^{xy}}{2y+xe^{xy}}\) wherever \(2y+xe^{xy}\ne0\).

- Use the chain rule for \(\tan(\sin x)\). - Replace \(\sec^2(\sin x)\) with \(1+\tan^2(\sin x)\). - State the conclusion only on intervals where the tangent expression is defined.

1. Differentiate the candidate: \(y'=\sec^2(\sin x)\cos x=(1+y^2)\cos x\). 2. The derivative equals the right side of \(y'=(1+y^2)\cos x\) wherever the candidate is defined. 3. Therefore, the candidate is a solution.

Yes; the candidate satisfies the differential equation on its domain.

- Differentiate the proposed function carefully. - Simplify the right side only where its denominator is nonzero. - A solution interval must lie within the domain of the differential equation.

1. Differentiate the candidate: \(y'=5\cos x\). 2. Where \(\sin x\ne 0\), \(y\cot x=5\sin x\cdot\frac{\cos x}{\sin x}=5\cos x\). 3. The differential equation is undefined at \(x=n\pi\), where \(n\) is an integer. 4. Therefore, the candidate is a solution on every interval \((n\pi,(n+1)\pi)\).

The candidate is a solution on every interval \((n\pi,(n+1)\pi)\), where \(n\) is an integer; it is not a solution on all real numbers because \(\cot x\) is undefined at \(x=n\pi\).

- Differentiate \(\ln(e^x+2)\) with the chain rule. - Rewrite \(e^{x-y}\) as \(e^x/e^y\). - Use the candidate relation \(e^y=e^x+2\) to compare both sides.

1. Differentiate the candidate: \(y'=\frac{e^x}{e^x+2}=e^{x-y}\). 2. The derivative equals the right side of \(y'=e^{x-y}\) wherever the candidate is defined. 3. Therefore, the candidate is a solution.

Yes; the candidate satisfies the differential equation on its domain.

- Let \(u=\frac{1}{1+e^{-x}}\), so the candidate is \(y=u+1\). - Compare the sign of \(y'\) with the sign of \(y(1-y)\) before doing lengthy algebra. - A sign contradiction on the candidate’s domain proves it is not a solution.

1. Differentiate the candidate: \(y'=\frac{e^{-x}}{(1+e^{-x})^2}\). 2. Substitute the candidate into the right side: \(y(1-y)=-\frac{2+e^{-x}}{(1+e^{-x})^2}\). 3. The two expressions have opposite signs for every real \(x\), so the candidate is not a solution.

No; the derivative does not equal the required right side for all \(x\).

- Use \(\frac{d}{dx}\ln y=\frac{y'}{y}\), valid because \(y>0\). - Factor \(y'\) from the two derivative terms on the left. - Solve for \(y'\) without dropping the stated positivity condition.

1. Differentiate the relation implicitly: \(y'+\frac{y'}{y}=1\). 2. Solve for the derivative to obtain \(y'=\frac{y}{y+1}\). 3. This matches the differential equation at all points where its denominator is nonzero.

The relation verifies \(y'=\frac{y}{y+1}\) on each differentiable branch where the differential equation is defined.

- Differentiate \(\arctan y\) with the chain rule. - Multiply by \(1+y^2\) after isolating the derivative factor. - Compare the result with the stated right side on each differentiable branch.

1. Differentiate the relation implicitly: \(\frac{y'}{1+y^2}=2x\). 2. Solve for the derivative to obtain \(y'=2x(1+y^2)\). 3. This matches the differential equation at every point on each differentiable branch of the relation.

The relation verifies \(y'=2x(1+y^2)\) on each differentiable branch where the differential equation is defined.

- Differentiate \((x^2+c)^{-1}\) while treating \(c\) as constant. - Rewrite \(y^2\) with the same denominator power as the derivative. - Decide whether the identity restricts \(c\), then state where the candidate is defined.

1. Differentiate the candidate: \(y'=-\frac{2x}{(x^2+c)^2}=-2xy^2\). 2. The differential equation is satisfied independently of the value of \(c\), wherever \(x^2+c\ne 0\). 3. Thus \(c\) is arbitrary subject to the candidate being defined.

\(c\) may be any real constant for which the candidate is defined on the interval considered.

- Differentiate \((x+C)^n\) and keep both the coefficient and exponent. - Rewrite \(y^{2/3}\) in terms of \(x+C\) on an interval where the powers are defined. - Match both the exponents and the numerical coefficients.

1. Differentiate: \(y'=n(x+C)^{n-1}\). 2. The right side is \(3y^{2/3}=3(x+C)^{2n/3}\). 3. Matching coefficients and exponents gives \(n=3\).

\(n=3\)

- Differentiate \(\tan(cx)\) using the chain rule. - Use \(\sec^2(cx)=1+\tan^2(cx)=1+y^2\). - Match the remaining coefficient and then apply the condition \(c>0\).

1. Differentiate the candidate: \(y'=c\sec^2(cx)=c(1+y^2)\). 2. Substitute into \(y'=4(1+y^2)\) and compare coefficients. 3. The required value is \(c=4\).

\(c=4\)

- Differentiate the complete candidate, including its constant factor. - Apply the chain rule to the exponent \(x^2\). - Verification must hold throughout the relevant domain, not at a single point.

1. Apply the chain rule to the complete candidate: \(y'=2xCe^{x^2}=2xy\). 2. The required right side is \(xy\), so the two sides are not equal for the family in general. 3. They agree identically only when \(C=0\), which gives the zero solution.

Jordan omitted the factor \(2\) from the chain rule. In fact, \(y'=2xCe^{x^2}=2xy\), so the family does not solve \(y'=xy\) except when \(C=0\).

- Checking one point cannot verify an entire family of curves. - Differentiate \(x^2+y^2=C\) implicitly and solve for \(y'\). - State the restriction at points where the quotient \(-x/y\) is undefined.

1. Checking one point is insufficient to verify an entire family of curves. 2. Differentiate the relation implicitly: \(2x+2yy'=0\). 3. Solving gives \(y'=-x/y\), so each differentiable branch satisfies the equation at every point where \(y\ne0\).

Checking one point is insufficient. Implicit differentiation gives \(2x+2yy'=0\), hence \(y'=-x/y\) on each differentiable branch where \(y\ne0\).

- Write the absolute-value function separately on the positive and negative half-lines. - A piecewise verification is incomplete unless differentiability at the joining point is checked. - Use the derivative definition at the joining point.

1. For \(x>0\), \(y=x^3\), so \(y''=6x=6|x|\). For \(x<0\), \(y=-x^3\), so \(y''=-6x=6|x|\). 2. The first derivative can be written as \(y'=3x|x|\), including \(y'(0)=0\). 3. At \(x=0\), \(y''(0)=\lim_{h\to0}\frac{3h|h|-0}{h}=\lim_{h\to0}3|h|=0\). 4. Since \(6|0|=0\), the differential equation also holds at the joining point.

Yes. The function satisfies \(y''=6|x|\) for all real \(x\), including \(x=0\).

- View the candidate as a product involving a negative one-half power. - After differentiating, rewrite the term containing the arbitrary constant using \(y\) itself. - Check whether the square-root denominator can vanish for a real input.

1. Differentiate using the product rule: \(y'=\frac{1}{(1+x^2)^{3/2}}-\frac{x(C+\arctan x)}{(1+x^2)^{3/2}}\). 2. Since \(y=\frac{C+\arctan x}{\sqrt{1+x^2}}\), the second term is \(-\frac{xy}{1+x^2}\). 3. Thus \((1+x^2)y'+xy=\frac{1}{\sqrt{1+x^2}}-xy+xy=\frac{1}{\sqrt{1+x^2}}\). 4. The family is defined for every real \(x\).

Yes. Every member of the family satisfies the differential equation on \((-\infty,\infty)\).

- Use the chain rule on the exponential whose exponent is a product. - Differentiate the product in the exponent implicitly. - Collect the terms containing \(y'\) before dividing.

1. Differentiate the relation implicitly: \(e^{xy}(y+xy')+1=0\). 2. Rearranging gives \(xe^{xy}y'=-(ye^{xy}+1)\). 3. Where \(xe^{xy}\ne0\), which requires \(x\ne0\), \(y'=-\frac{ye^{xy}+1}{xe^{xy}}\).

Yes. Each differentiable branch satisfies the differential equation at points where \(x\ne0\).

- Name the parenthetical expression before differentiating. - Use both the product rule and the Fundamental Theorem of Calculus. - Replace the repeated product by \(y\) before simplifying.

1. Let \(F(x)=C+\int_0^x e^{t^2}\sin t\,dt\), so \(F'(x)=e^{x^2}\sin x\). 2. Differentiate \(y=e^{-x^2}F(x)\): \(y'=-2xe^{-x^2}F(x)+e^{-x^2}F'(x)\). 3. Since \(e^{-x^2}F(x)=y\), this becomes \(y'=-2xy+\sin x\). 4. Therefore, \(y'+2xy=\sin x\).

The family satisfies \(y'+2xy=\sin x\) for every real \(x\).

- Differentiate \((x+C)^2\) directly. - Simplify \(\sqrt{(x+C)^2}\) as an absolute value, not automatically as \(x+C\). - Determine on which side of \(x=-C\) the derivative matches the right side.

1. Identify the derivative order and differentiate the complete candidate or relation. 2. The derivative is \(2(x+C)\), while \(2\sqrt y=2|x+C|\). They agree only on intervals where \(x+C\ge 0\).

The derivative is \(2(x+C)\), while \(2\sqrt y=2|x+C|\). They agree only on intervals where \(x+C\ge 0\).