Verifying Solutions - Differential Equations

1. The function $y(x) = a_0(1-3x^2)$ is a polynomial solution to Legendre's equation for order 2, $(1-x^2)y'' - 2xy' + 6y = 0$. For what value of the constant $a_0$ does this solution satisfy the condition $y'(1)=12$?
   Answer: $a_0 = -2$ The problem requires finding the value of $a_0$ that satisfies the given condition. While the stem states that $y(x) = a_0(1-3x^2)$ is a solution, the core task is to use the condition on the derivative. First, find the derivative of the solution: $y(x) = a_0 - 3a_0x^2$ $y'(x) = -6a_0x$ Next, apply the condition $y'(1)=12$: $y'(1) = -6a_0(1) = -6a_0$ Set this equal to 12: $-6a_0 = 12$ $a_0 = -2$ A common mistake is to apply the condition to $y(x)$ instead of $y'(x)$, which would lead to $y(1) = a_0(1-3(1)^2) = -2a_0 = 12$, yielding $a_0=-6$. Another error is a sign mistake, yielding $a_0=2$.
2. Consider the separable differential equation $\frac{dy}{dx} = \frac{y\ln(y)}{x}$ with $y > 1$. A student claims the general solution is $\ln(\ln(y)) = \ln|x| + C$. To verify this solution, what is the correct approach and conclusion?
   Answer: Differentiate implicitly: $\frac{1}{\ln(y)} \cdot \frac{1}{y} \cdot \frac{dy}{dx} = \frac{1}{x}$, so $\frac{dy}{dx} = \frac{y\ln(y)}{x}$. The solution is verified as correct. To verify the implicit solution $\ln(\ln(y)) = \ln|x| + C$, we differentiate both sides with respect to $x$. Using the chain rule on the left side: $\frac{d}{dx}[\ln(\ln(y))] = \frac{1}{\ln(y)} \cdot \frac{d}{dx}[\ln(y)] = \frac{1}{\ln(y)} \cdot \frac{1}{y} \cdot \frac{dy}{dx}$. The right side gives: $\frac{d}{dx}[\ln|x| + C] = \frac{1}{x}$. Therefore: $\frac{1}{\ln(y)} \cdot \frac{1}{y} \cdot \frac{dy}{dx} = \frac{1}{x}$. Solving for $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{y\ln(y)}{x}$, which exactly matches the original differential equation. Choice B omits the $\frac{1}{y}$ factor from the chain rule. Choice C describes deriving the solution rather than verifying it. Choice D focuses on a domain issue that doesn't affect the verification process.
3. Consider the differential equation $\frac{dy}{dx} = \frac{2xy}{x^2 + 1}$. A student claims that $y = C(x^2 + 1)$ is the general solution, where $C$ is an arbitrary constant. To verify this solution, which of the following steps correctly identifies whether the student's claim is valid?
   Answer: Substitute into the original equation: $\frac{d}{dx}[C(x^2 + 1)] = 2Cx$ and $\frac{2xy}{x^2 + 1} = \frac{2x \cdot C(x^2 + 1)}{x^2 + 1} = 2Cx$. Since both sides equal $2Cx$, the solution is correct. To verify a solution, we substitute it into the original differential equation. Computing $\frac{d}{dx}[C(x^2 + 1)] = C \cdot 2x = 2Cx$. Substituting $y = C(x^2 + 1)$ into the right side: $\frac{2xy}{x^2 + 1} = \frac{2x \cdot C(x^2 + 1)}{x^2 + 1} = 2Cx$. Since both sides equal $2Cx$, the solution is verified. Choice B incorrectly states that $C(2x) \neq 2Cx$, which is false since multiplication is commutative. Choice C makes an error about the constant—when we have $\ln|y| = \ln(x^2 + 1) + C$, this gives $y = \pm e^C(x^2 + 1)$, and $\pm e^C$ can represent any nonzero constant. Choice D is too superficial and doesn't actually perform the verification.
4. If the function $y(x) = C_1 e^{-2x} + Bx + C$ is a family of solutions to the non-homogeneous differential equation $y' + 2y = 4x$, what must be the values of the constants $B$ and $C$?
   Answer: $B = 2, C = -1$ We need to find the particular solution part, which is $y_p = Bx+C$. Its derivative is $y_p' = B$. Substitute these into the differential equation: $y' + 2y = 4x$ $(B) + 2(Bx+C) = 4x$ $2Bx + (B+2C) = 4x$ For this equation to hold for all $x$, we must equate the coefficients of the powers of $x$ on both sides. Coefficient of $x$: $2B = 4 \implies B = 2$. Constant term: $B+2C = 0$. Substituting $B=2$, we get $2+2C = 0 \implies 2C = -2 \implies C = -1$. Thus, $B=2$ and $C=-1$.
5. Which of the following functions, defined by an integral, is a solution to the initial value problem $y' + 2xy = 1$, with the initial condition $y(0)=0$?
   Answer: $y(x) = e^{-x^2} \int_0^x e^{t^2} dt$ We must check which function satisfies both the differential equation and the initial condition. Let's verify choice A: $y(x) = e^{-x^2} \int_0^x e^{t^2} dt$. First, check the initial condition: $y(0) = e^{-0^2} \int_0^0 e^{t^2} dt = 1 \cdot 0 = 0$. The condition is met. Next, find the derivative using the product rule and the Fundamental Theorem of Calculus: $y'(x) = \frac{d}{dx}(e^{-x^2}) \cdot \int_0^x e^{t^2} dt + e^{-x^2} \cdot \frac{d}{dx}(\int_0^x e^{t^2} dt)$ $y'(x) = (-2xe^{-x^2}) \int_0^x e^{t^2} dt + e^{-x^2} \cdot e^{x^2}$ $y'(x) = -2x \left(e^{-x^2} \int_0^x e^{t^2} dt\right) + 1$ Recognize that the term in the parenthesis is the original function $y(x)$. So, $y'(x) = -2xy(x) + 1$. Rearranging gives $y' + 2xy = 1$. The differential equation is satisfied. Choice C is a notational variant of A, so it is also correct, but typically only one such form would appear. Let's treat A as the intended answer. The other choices fail the differentiation step.
6. For what value of the constant $k$ is the function $u(x,t) = e^{-9kt}\sin(3x)$ a solution to the one-dimensional heat equation $u_t = u_{xx}$?
   Answer: $k = 1$ We need to compute the partial derivatives $u_t$ and $u_{xx}$ and substitute them into the heat equation. For $u(x,t) = e^{-9kt}\sin(3x)$: Partial derivative with respect to $t$: $u_t = \frac{\partial}{\partial t} (e^{-9kt}\sin(3x)) = -9k e^{-9kt}\sin(3x)$. Partial derivatives with respect to $x$: $u_x = \frac{\partial}{\partial x} (e^{-9kt}\sin(3x)) = 3e^{-9kt}\cos(3x)$. $u_{xx} = \frac{\partial}{\partial x} (3e^{-9kt}\cos(3x)) = -9e^{-9kt}\sin(3x)$. Now substitute into $u_t = u_{xx}$: $-9k e^{-9kt}\sin(3x) = -9e^{-9kt}\sin(3x)$ Dividing both sides by the non-zero term $-9e^{-9kt}\sin(3x)$ yields $k=1$.
7. For the function $y(x) = (Ax + B)e^{kx}$ to be the general solution to the differential equation $y'' - 4y' + 4y = 0$, what must be the value of the constant $k$?
   Answer: $k = 2$ To verify the solution, we must compute the first and second derivatives of $y(x)$ and substitute them into the differential equation. First derivative: $y' = A e^{kx} + k(Ax+B)e^{kx} = (A + kAx + kB)e^{kx}$. Second derivative: $y'' = kAe^{kx} + k(A + kAx + kB)e^{kx} = (2kA + k^2Ax + k^2B)e^{kx}$. Substitute into the equation: $(2kA + k^2Ax + k^2B)e^{kx} - 4(A + kAx + kB)e^{kx} + 4(Ax + B)e^{kx} = 0$ Divide by $e^{kx}$ (which is never zero) and group terms by powers of $x$: $x(k^2A - 4kA + 4A) + (2kA + k^2B - 4A - 4kB + 4B) = 0$ For this to hold for all $x$ and for arbitrary constants $A$ and $B$, the coefficients of the powers of $x$ must be zero. The coefficient of $x$ is $A(k^2 - 4k + 4)$. Since this must be true for any $A$, we must have $k^2 - 4k + 4 = 0$, which factors as $(k-2)^2 = 0$. The only solution is $k=2$. Checking the constant term with $k=2$ also yields zero.
8. Consider the boundary value problem $y'' + 4y = 0$ with $y(0) = 2$ and $y\left(\frac{\pi}{4}\right) = 1$. A student proposes the solution $y = 2\cos(2x) + \sin(2x)$. What is the most efficient way to determine if this solution is correct?
   Answer: Check that $y'' = -8\cos(2x) - 4\sin(2x)$ and verify $y'' + 4y = -8\cos(2x) - 4\sin(2x) + 8\cos(2x) + 4\sin(2x) = 0$, then confirm boundary conditions. When verifying a proposed solution to a boundary value problem, you need to check both that it satisfies the differential equation and meets the boundary conditions. The most efficient approach depends on quickly identifying any failures. Let's examine the proposed solution $y = 2\cos(2x) + \sin(2x)$. First, we find the second derivative: $y' = -4\sin(2x) + 2\cos(2x)$ and $y'' = -8\cos(2x) - 4\sin(2x)$. Substituting into the differential equation: $y'' + 4y = -8\cos(2x) - 4\sin(2x) + 4(2\cos(2x) + \sin(2x)) = -8\cos(2x) - 4\sin(2x) + 8\cos(2x) + 4\sin(2x) = 0$. The differential equation is satisfied, so we should check boundary conditions. Choice A makes a calculation error, claiming $y'' + 4y = -4\cos(2x)$, but this ignores the $4y$ term properly. Choice B contains a dangerous assumption—while $\cos(2x)$ and $\sin(2x)$ are indeed solutions to $y'' + 4y = 0$, you can't assume any linear combination works without verifying the coefficients don't introduce errors in your specific proposal. Choice D describes solving the problem from scratch rather than verifying the given solution, which is inefficient when you're asked to check a specific proposal. Choice C correctly shows the complete verification: computing $y''$, demonstrating the differential equation is satisfied, then checking boundary conditions. Strategy tip: When verifying solutions, always check the differential equation first with careful algebra—calculation errors here can save you time by avoiding unnecessary boundary condition work.
9. A student solving $\frac{dy}{dx} + \frac{2y}{x} = x^2$ claims the solution is $y = \frac{x^4}{6} + \frac{C}{x^2}$. To verify this solution completely, which approach correctly identifies all necessary checks?
   Answer: Compute $y' = \frac{2x^3}{3} - \frac{2C}{x^3}$, then substitute: $\frac{2x^3}{3} - \frac{2C}{x^3} + \frac{2}{x} \cdot \left(\frac{x^4}{6} + \frac{C}{x^2}\right) = \frac{2x^3}{3} - \frac{2C}{x^3} + \frac{x^3}{3} + \frac{2C}{x^3} = x^3 \neq x^2$. To verify the solution $y = \frac{x^4}{6} + \frac{C}{x^2}$, we compute $y' = \frac{4x^3}{6} - \frac{2C}{x^3} = \frac{2x^3}{3} - \frac{2C}{x^3}$. Substituting into $\frac{dy}{dx} + \frac{2y}{x} = x^2$: $\frac{2x^3}{3} - \frac{2C}{x^3} + \frac{2}{x}\left(\frac{x^4}{6} + \frac{C}{x^2}\right) = \frac{2x^3}{3} - \frac{2C}{x^3} + \frac{2x^4}{6x} + \frac{2C}{x^3} = \frac{2x^3}{3} - \frac{2C}{x^3} + \frac{x^3}{3} + \frac{2C}{x^3} = \frac{3x^3}{3} = x^3$. Since $x^3 \neq x^2$, the solution is incorrect. Choice A has computational errors in the substitution. Choice C makes the same correct calculation but incorrectly concludes the solution is right. Choice D suggests an incomplete verification approach that doesn't actually check if the proposed solution works.
10. The differential equation $x^2y'' - 2xy' + 2y = 0$ has a proposed solution $y = x^r$ for some constant $r$. After substituting $y = x^r$ into the equation and simplifying, which condition on $r$ must be satisfied?
    Answer: $r(r-1) - 2r + 2 = 0$, which simplifies to $r^2 - 3r + 2 = 0$, giving $r = 1$ or $r = 2$. For $y = x^r$, we compute $y' = rx^{r-1}$ and $y'' = r(r-1)x^{r-2}$. Substituting into $x^2y'' - 2xy' + 2y = 0$: $x^2 \cdot r(r-1)x^{r-2} - 2x \cdot rx^{r-1} + 2x^r = 0$. This simplifies to $r(r-1)x^r - 2rx^r + 2x^r = 0$. Factoring out $x^r$ (assuming $x \neq 0$): $x^r[r(r-1) - 2r + 2] = 0$. Thus we need $r(r-1) - 2r + 2 = 0$, which gives $r^2 - r - 2r + 2 = r^2 - 3r + 2 = 0$. Factoring: $(r-1)(r-2) = 0$, so $r = 1$ or $r = 2$. Choice B uses wrong derivative formula. Choice C skips steps and gets wrong equation. Choice D correctly sets up the equation but falsely claims no real solutions when the discriminant is $9 - 8 = 1 > 0$.
11. A student solving the initial value problem $y'' - 5y' + 6y = 0$ with $y(0) = 1$ and $y'(0) = 4$ obtains $y = 2e^{2x} - e^{3x}$. To verify this solution, which computation reveals whether the student's work is correct?
    Answer: $y' = 4e^{2x} - 3e^{3x}$ and $y'' = 8e^{2x} - 9e^{3x}$. Checking the DE: $8e^{2x} - 9e^{3x} - 5(4e^{2x} - 3e^{3x}) + 6(2e^{2x} - e^{3x}) = 0$, and $y(0) = 1$, $y'(0) = 1 \neq 4$. To fully verify the solution, we need to check both the differential equation and initial conditions. Computing derivatives: $y' = 4e^{2x} - 3e^{3x}$ and $y'' = 8e^{2x} - 9e^{3x}$. Checking the DE: $y'' - 5y' + 6y = (8e^{2x} - 9e^{3x}) - 5(4e^{2x} - 3e^{3x}) + 6(2e^{2x} - e^{3x}) = 8e^{2x} - 9e^{3x} - 20e^{2x} + 15e^{3x} + 12e^{2x} - 6e^{3x} = 0e^{2x} + 0e^{3x} = 0$ ✓. Checking initial conditions: $y(0) = 2e^0 - e^0 = 2 - 1 = 1$ ✓, but $y'(0) = 4e^0 - 3e^0 = 4 - 3 = 1 \neq 4$ ✗. While choices A, B, and C all correctly identify that $y'(0) = 1 \neq 4$, choice D provides the complete verification by checking both the differential equation and initial conditions. Choices A and B contain minor notational issues in their derivative calculations.
12. A student performs a substitution to check if $y(x) = \sin(x) + x$ is a solution to the differential equation $y'' + y = x$. The student's work is shown below. 1. $y = \sin(x) + x$ 2. $y' = \cos(x) + 1$ 3. $y'' = -\sin(x) + 1$ 4. Substitute into $y'' + y$: $(-\sin(x) + 1) + (\sin(x) + x)$ 5. Simplify: $(-\sin(x) + \sin(x)) + 1 + x = 1+x$ 6. Conclusion: The function is not a solution because $1+x \neq x$. In which step did the student first make an error?
    Answer: Step 3: Incorrect differentiation of $y' $. Let's review the student's steps. Step 1: The given function is $y = \sin(x) + x$. This is correct. Step 2: The first derivative is $y' = \frac{d}{dx}(\sin(x) + x) = \cos(x) + 1$. This is correct. Step 3: The second derivative is $y'' = \frac{d}{dx}(\cos(x) + 1) = -\sin(x) + 0 = -\sin(x)$. The student wrote $y'' = -\sin(x) + 1$, which is incorrect. The derivative of the constant 1 is 0. This is the first error. If the calculation were correct, $y'' = -\sin(x)$. Then $y''+y = -\sin(x) + (\sin(x)+x) = x$. The function is actually a solution, and the student's incorrect conclusion in Step 6 stems from the error in Step 3.
13. For what positive value of the constant $\omega$ is the function $y(t) = C e^{-3t}\sin(\omega t)$ a solution to the differential equation $y'' + 6y' + 13y = 0$ for some non-zero constant $C$?
    Answer: $\omega = 2$ A standard approach is to find the roots of the characteristic equation $r^2 + 6r + 13 = 0$. Using the quadratic formula: $r = \frac{-6 \pm \sqrt{6^2 - 4(1)(13)}}{2} = \frac{-6 \pm \sqrt{36 - 52}}{2} = \frac{-6 \pm \sqrt{-16}}{2} = \frac{-6 \pm 4i}{2} = -3 \pm 2i$. The general solution for these complex conjugate roots is $y(t) = e^{-3t}(C_1 \cos(2t) + C_2 \sin(2t))$. The proposed function $y(t) = C e^{-3t}\sin(\omega t)$ matches this form if $\omega = 2$. Alternatively, one can compute $y'$ and $y''$ for $y = e^{-3t}\sin(\omega t)$ and substitute them into the DE to solve for $\omega$, but this is more labor-intensive.
14. The implicit relation $x^2 - 2xy - y^2 = C$, where $C$ is an arbitrary constant, defines a family of solutions to which of the following first-order differential equations?
    Answer: $y' = \frac{x-y}{x+y}$ We use implicit differentiation with respect to $x$ on the given relation $x^2 - 2xy - y^2 = C$. $\frac{d}{dx}(x^2 - 2xy - y^2) = \frac{d}{dx}(C)$ Applying the product rule to the $-2xy$ term and the chain rule to the $-y^2$ term, we get: $2x - (2y \cdot 1 + 2x \cdot y') - 2y \cdot y' = 0$ $2x - 2y - 2xy' - 2yy' = 0$ Now, we solve for $y'$. $2x - 2y = 2xy' + 2yy'$ $2(x - y) = 2(x + y)y'$ $y' = \frac{x-y}{x+y}$ Distractors are based on common sign errors during differentiation or algebraic manipulation.
15. The differential equation $x^2 y'' - 3xy' + 4y = 0$ has a solution of the form $y = x^k$ for $x > 0$. Which of the following describes all possible real values of $k$?
    Answer: $k = 2$ only To check for a solution of the form $y=x^k$, we compute its derivatives: $y' = kx^{k-1}$ and $y'' = k(k-1)x^{k-2}$. We substitute these into the Cauchy-Euler equation: $x^2(k(k-1)x^{k-2}) - 3x(kx^{k-1}) + 4(x^k) = 0$ $k(k-1)x^k - 3kx^k + 4x^k = 0$ Factor out $x^k$ (which is non-zero for $x>0$): $x^k(k(k-1) - 3k + 4) = 0$ For this to be true, the expression in the parenthesis must be zero. This gives the auxiliary equation: $k^2 - k - 3k + 4 = 0$ $k^2 - 4k + 4 = 0$ $(k-2)^2 = 0$ This equation has a single, repeated root at $k=2$.
16. A function of the form $\mathbf{x}(t) = \begin{pmatrix} e^{3t} \\ k e^{3t} \end{pmatrix}$ is a solution to the system of differential equations $\dot{x} = x+y$, $\dot{y} = 4x+y$. What is the value of the constant $k$?
    Answer: $k = 2$ Let $x(t) = e^{3t}$ and $y(t) = ke^{3t}$. We find their derivatives: $\dot{x} = 3e^{3t}$ and $\dot{y} = 3ke^{3t}$. Now, we substitute these into the system of equations. First equation: $\dot{x} = x+y$ becomes $3e^{3t} = e^{3t} + ke^{3t}$. Dividing by $e^{3t}$ gives $3 = 1+k$, which implies $k=2$. Second equation: $\dot{y} = 4x+y$ becomes $3ke^{3t} = 4e^{3t} + ke^{3t}$. Dividing by $e^{3t}$ gives $3k = 4+k$, which implies $2k=4$, so $k=2$. Both equations require $k=2$. The distractor $k=-2$ corresponds to the other eigenvector for this system.
17. For what value(s) of the constant $k$ is the function $y(x) = \cosh(x) + k\sin(x)$ a solution to the differential equation $y^{(4)} - y = 0$?
    Answer: For all real values of $k$ We need to compute the fourth derivative of $y(x)$. $y(x) = \cosh(x) + k\sin(x)$ $y'(x) = \sinh(x) + k\cos(x)$ $y''(x) = \cosh(x) - k\sin(x)$ $y'''(x) = \sinh(x) - k\cos(x)$ $y^{(4)}(x) = \cosh(x) + k\sin(x)$ We see that $y^{(4)}(x)$ is identical to the original function $y(x)$. Therefore, substituting into the differential equation gives: $y^{(4)} - y = (\cosh(x) + k\sin(x)) - (\cosh(x) + k\sin(x)) = 0$. This equation is true for any real value of the constant $k$. The question seems to imply a unique value for $k$ must be found, but the verification shows it holds for all $k$.
18. For the differential equation $\frac{dy}{dx} = \frac{x-y}{x+y}$, a student claims that $x^2 - 2xy - y^2 = C$ is an implicit solution. Which verification approach correctly determines the validity of this claim?
    Answer: Differentiate implicitly to get $2x - 2y - 2x\frac{dy}{dx} - 2y\frac{dy}{dx} = 0$, solve for $\frac{dy}{dx} = \frac{2x-2y}{2x+2y} = \frac{x-y}{x+y}$, confirming the solution is correct. When verifying whether a proposed function is a solution to a differential equation, you need to differentiate the implicit solution and check if it produces the original equation. This requires careful application of implicit differentiation rules. To verify $x^2 - 2xy - y^2 = C$, you differentiate both sides with respect to $x$. The left side becomes: $2x - 2y - 2x\frac{dy}{dx} - 2y\frac{dy}{dx} = 0$. Notice that the $-2xy$ term produces $-2y - 2x\frac{dy}{dx}$ using the product rule, and $-y^2$ gives $-2y\frac{dy}{dx}$. Factoring out the $\frac{dy}{dx}$ terms: $2x - 2y - 2(x + y)\frac{dy}{dx} = 0$. Solving for $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{2x - 2y}{2(x + y)} = \frac{x - y}{x + y}$, which matches the original differential equation exactly. Option A contains the correct differentiation but unnecessarily emphasizes the domain restriction $x + y \neq 0$. While mathematically true, this doesn't affect the verification process itself. Option B makes an error in the implicit differentiation of the $-2xy$ term, writing $-2x\frac{dy}{dx}$ instead of the correct $-2y - 2x\frac{dy}{dx}$. Option D attempts verification through specific values, which can never confirm a general solution—you need algebraic verification. Study tip: When verifying implicit solutions, always differentiate the entire proposed solution term-by-term using the chain rule and product rule. Substituting specific values only works for checking particular solutions, not general ones.

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