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Verifying DE Solutions - Calculus 2

  1. The function y=Ce−x2/2y = Ce^{-x^2/2} is proposed as a solution to the differential equation y′+xy=0y' + xy = 0. However, when checking the initial condition y(1)=3y(1) = 3, which of the following statements about the verification process is correct?
    Answer: The function satisfies the differential equation and the initial condition with C=3e1/2C = 3e^{1/2}, providing a complete solution First, verify the differential equation: y=Ce−x2/2y = Ce^{-x^2/2}, so y′=C⋅(−x)e−x2/2=−Cxe−x2/2y' = C \cdot (-x)e^{-x^2/2} = -Cxe^{-x^2/2}. Substituting into y′+xy=0y' + xy = 0: −Cxe−x2/2+x(Ce−x2/2)=−Cxe−x2/2+Cxe−x2/2=0-Cxe^{-x^2/2} + x(Ce^{-x^2/2}) = -Cxe^{-x^2/2} + Cxe^{-x^2/2} = 0 ✓. For the initial condition y(1)=3y(1) = 3: Ce−1/2=3Ce^{-1/2} = 3, so C=3e1/2C = 3e^{1/2}. Therefore, y=3e1/2e−x2/2=3e(1−x2)/2y = 3e^{1/2}e^{-x^2/2} = 3e^{(1-x^2)/2}. Checking: y(1)=3e(1−1)/2=3e0=3y(1) = 3e^{(1-1)/2} = 3e^0 = 3 ✓. Choice A incorrectly states the initial condition fails. Choice C incorrectly claims the differential equation isn't satisfied. Choice D has the wrong value for C.
  2. For the differential equation d2ydx2+4dydx+4y=0\frac{d^2y}{dx^2} + 4\frac{dy}{dx} + 4y = 0, a student proposes the solution y=(Ax+B)e−2xy = (Ax + B)e^{-2x}. When verifying this solution, what should be the result of computing y′′+4y′+4yy'' + 4y' + 4y?
    Answer: The expression simplifies to 00, confirming the proposed solution is correct Given y=(Ax+B)e−2xy = (Ax + B)e^{-2x}, we compute the derivatives. Using the product rule: y′=Ae−2x+(Ax+B)(−2)e−2x=Ae−2x−2(Ax+B)e−2x=[A−2(Ax+B)]e−2x=(A−2Ax−2B)e−2xy' = Ae^{-2x} + (Ax + B)(-2)e^{-2x} = Ae^{-2x} - 2(Ax + B)e^{-2x} = [A - 2(Ax + B)]e^{-2x} = (A - 2Ax - 2B)e^{-2x}. For the second derivative: y′′=(−2A)e−2x+(A−2Ax−2B)(−2)e−2x=−2Ae−2x−2(A−2Ax−2B)e−2x=[−2A−2A+4Ax+4B]e−2x=(−4A+4Ax+4B)e−2xy'' = (-2A)e^{-2x} + (A - 2Ax - 2B)(-2)e^{-2x} = -2Ae^{-2x} - 2(A - 2Ax - 2B)e^{-2x} = [-2A - 2A + 4Ax + 4B]e^{-2x} = (-4A + 4Ax + 4B)e^{-2x}. Now substituting into y′′+4y′+4yy'' + 4y' + 4y: (−4A+4Ax+4B)e−2x+4(A−2Ax−2B)e−2x+4(Ax+B)e−2x=e−2x[(−4A+4Ax+4B)+4(A−2Ax−2B)+4(Ax+B)]=e−2x[−4A+4Ax+4B+4A−8Ax−8B+4Ax+4B]=e−2x[(−4A+4A)+(4Ax−8Ax+4Ax)+(4B−8B+4B)]=e−2x[0+0+0]=0(-4A + 4Ax + 4B)e^{-2x} + 4(A - 2Ax - 2B)e^{-2x} + 4(Ax + B)e^{-2x} = e^{-2x}[(-4A + 4Ax + 4B) + 4(A - 2Ax - 2B) + 4(Ax + B)] = e^{-2x}[-4A + 4Ax + 4B + 4A - 8Ax - 8B + 4Ax + 4B] = e^{-2x}[(-4A + 4A) + (4Ax - 8Ax + 4Ax) + (4B - 8B + 4B)] = e^{-2x}[0 + 0 + 0] = 0. The other choices represent common computational errors in the differentiation process.
  3. A function y(x)y(x) satisfies the differential equation xy′−2y=x3exxy' - 2y = x^3e^x. To verify that y=x2(ex+C)y = x^2(e^x + C) is indeed a solution, which step in the verification process would reveal an error if one exists?
    Answer: Verifying that xy′−2y=x(2x(ex+C)+x2ex)−2x2(ex+C)xy' - 2y = x(2x(e^x + C) + x^2e^x) - 2x^2(e^x + C) simplifies to x3exx^3e^x To verify the solution, we need to substitute into the differential equation and check if both sides are equal. First, y=x2(ex+C)=x2ex+Cx2y = x^2(e^x + C) = x^2e^x + Cx^2. Then y′=ddx[x2ex+Cx2]=2xex+x2ex+2Cx=2x(ex+C)+x2exy' = \frac{d}{dx}[x^2e^x + Cx^2] = 2xe^x + x^2e^x + 2Cx = 2x(e^x + C) + x^2e^x. Substituting into xy′−2yxy' - 2y: x[2x(ex+C)+x2ex]−2[x2(ex+C)]=2x2(ex+C)+x3ex−2x2(ex+C)=x3exx[2x(e^x + C) + x^2e^x] - 2[x^2(e^x + C)] = 2x^2(e^x + C) + x^3e^x - 2x^2(e^x + C) = x^3e^x. This confirms the solution is correct. Choice A only computes the derivative but doesn't complete the verification. Choice B checks form but not the actual equation. Choice D checks only the homogeneous part. Choice C is the complete verification step that would reveal any errors.
  4. Consider the initial value problem dydx=2y+x\frac{dy}{dx} = 2y + x, y(0)=1y(0) = 1. A student proposes the solution y=32e2x−x2−14y = \frac{3}{2}e^{2x} - \frac{x}{2} - \frac{1}{4}. To complete the verification, which condition must be checked last?
    Answer: That y(0)=32−0−14=54≠1y(0) = \frac{3}{2} - 0 - \frac{1}{4} = \frac{5}{4} \neq 1, showing the solution fails the initial condition When verifying a solution to an initial value problem, we must check both that the function satisfies the differential equation AND the initial condition. The verification should proceed by: (1) checking that the proposed solution satisfies the differential equation, and (2) checking that it satisfies the initial condition. For the proposed solution y=32e2x−x2−14y = \frac{3}{2}e^{2x} - \frac{x}{2} - \frac{1}{4}, we have y′=3e2x−12y' = 3e^{2x} - \frac{1}{2}. The right side gives 2y+x=2(32e2x−x2−14)+x=3e2x−x−12+x=3e2x−122y + x = 2\left(\frac{3}{2}e^{2x} - \frac{x}{2} - \frac{1}{4}\right) + x = 3e^{2x} - x - \frac{1}{2} + x = 3e^{2x} - \frac{1}{2}, which equals y′y', so the differential equation is satisfied. However, checking the initial condition: y(0)=32e0−02−14=32−14=54≠1y(0) = \frac{3}{2}e^{0} - \frac{0}{2} - \frac{1}{4} = \frac{3}{2} - \frac{1}{4} = \frac{5}{4} \neq 1. Therefore, the proposed solution fails the initial condition. Choice B should be checked first, not last. Choices C and D involve finding the correct constant, which isn't what the question asks for.
  5. For the second-order linear differential equation y′′−3y′+2y=exy'' - 3y' + 2y = e^x, consider the proposed solution y=Aex+Be2x+12xexy = Ae^x + Be^{2x} + \frac{1}{2}xe^x. When verifying this solution, what is the most likely source of error in a student's work?
    Answer: Incorrectly computing d2dx2(12xex)=12ex\frac{d^2}{dx^2}\left(\frac{1}{2}xe^x\right) = \frac{1}{2}e^x instead of 12ex+xex\frac{1}{2}e^x + xe^x To verify the solution, we need to compute all derivatives correctly. For 12xex\frac{1}{2}xe^x: First derivative: ddx(12xex)=12(ex+xex)=12ex(1+x)\frac{d}{dx}\left(\frac{1}{2}xe^x\right) = \frac{1}{2}(e^x + xe^x) = \frac{1}{2}e^x(1 + x). Second derivative: d2dx2(12xex)=12ddx[ex+xex]=12[ex+ex+xex]=12ex(2+x)\frac{d^2}{dx^2}\left(\frac{1}{2}xe^x\right) = \frac{1}{2}\frac{d}{dx}[e^x + xe^x] = \frac{1}{2}[e^x + e^x + xe^x] = \frac{1}{2}e^x(2 + x). A common error is computing the second derivative as just 12ex\frac{1}{2}e^x by incorrectly applying the product rule or forgetting terms. The correct second derivative is 12ex(2+x)\frac{1}{2}e^x(2 + x), not 12ex\frac{1}{2}e^x. Choice B is incorrect because recognizing the homogeneous solution is usually straightforward. Choice C misunderstands how particular solutions work. Choice D describes an error in the first derivative, but the more critical error typically occurs with the second derivative since it involves more steps.
  6. A student is verifying that y=2xx2+1y = \frac{2x}{x^2 + 1} satisfies the differential equation dydx+2xyx2+1=2x2+1\frac{dy}{dx} + \frac{2xy}{x^2 + 1} = \frac{2}{x^2 + 1}. After computing dydx=2(1−x2)(x2+1)2\frac{dy}{dx} = \frac{2(1-x^2)}{(x^2+1)^2} using the quotient rule, what should their next step reveal?
    Answer: Substituting gives 2(1−x2)(x2+1)2+4x2(x2+1)2=2x2+1\frac{2(1-x^2)}{(x^2+1)^2} + \frac{4x^2}{(x^2+1)^2} = \frac{2}{x^2+1}, confirming the solution is correct Given y=2xx2+1y = \frac{2x}{x^2 + 1} and dydx=2(1−x2)(x2+1)2\frac{dy}{dx} = \frac{2(1-x^2)}{(x^2+1)^2}, we substitute into the left side of the differential equation. The term 2xyx2+1=2x⋅2xx2+1x2+1=4x2(x2+1)2\frac{2xy}{x^2 + 1} = \frac{2x \cdot \frac{2x}{x^2 + 1}}{x^2 + 1} = \frac{4x^2}{(x^2 + 1)^2}. Therefore: dydx+2xyx2+1=2(1−x2)(x2+1)2+4x2(x2+1)2=2(1−x2)+4x2(x2+1)2=2−2x2+4x2(x2+1)2=2+2x2(x2+1)2=2(1+x2)(x2+1)2=2x2+1\frac{dy}{dx} + \frac{2xy}{x^2 + 1} = \frac{2(1-x^2)}{(x^2+1)^2} + \frac{4x^2}{(x^2+1)^2} = \frac{2(1-x^2) + 4x^2}{(x^2+1)^2} = \frac{2 - 2x^2 + 4x^2}{(x^2+1)^2} = \frac{2 + 2x^2}{(x^2+1)^2} = \frac{2(1 + x^2)}{(x^2+1)^2} = \frac{2}{x^2+1}. This equals the right side, confirming the solution is correct. Choice A has the wrong final comparison. Choice B incorrectly simplifies the numerator. Choice C makes an error in the final equality. Choice D correctly shows the verification.

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