Differential Equations Quiz: Undetermined Coefficients

What this quiz covers

This quiz focuses on Undetermined Coefficients, giving you a quick way to practice the rules, question types, and explanations that matter most for Differential Equations.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. When $r$ is a repeated root of the characteristic equation $s^2 + as + b = 0$ (correct answer)
2. When $r$ is a simple root of the characteristic equation $s^2 + as + b = 0$
3. When $r^2 + ar + b = 0$ has complex conjugate roots involving $r$
4. When $r$ satisfies $r^2 + ar + b \neq 0$ but $2r + a = 0$

Explanation: When solving non-homogeneous linear differential equations with exponential forcing functions like $e^{rx}$, the form of your particular solution depends critically on whether the exponent $r$ relates to the roots of the characteristic equation $s^2 + as + b = 0$. The correct answer is A. When $r$ is a repeated root of the characteristic equation, it means $r$ satisfies $r^2 + ar + b = 0$ and appears twice in the homogeneous solution. Since both $e^{rx}$ and $xe^{rx}$ are already part of the homogeneous solution, you must multiply by $x^2$ to ensure your particular solution is linearly independent. This gives $y_p = Ax^2e^{rx}$. Choice B is incorrect because when $r$ is a simple root, only $e^{rx}$ appears in the homogeneous solution, so $y_p = Axe^{rx}$ (multiply by $x$ once) would be sufficient. Choice C misunderstands the relationship entirely. Complex conjugate roots don't directly determine the particular solution form for an $e^{rx}$ forcing function. Choice D describes an impossible scenario. If $r^2 + ar + b \neq 0$, then $r$ isn't a root, so the condition $2r + a = 0$ (which would make $r = -a/2$) is irrelevant to determining the particular solution form. Study tip: Always check if your forcing function's exponent matches roots of the characteristic equation. No match = try $Ae^{rx}$; simple root = try $Axe^{rx}$; repeated root = try $Ax^2e^{rx}$.

Question 2

1. $Y_p(t) = Ae^{3t}$
2. $Y_p(t) = Ate^{3t}$
3. $Y_p(t) = At^2e^{3t}$ (correct answer)
4. $Y_p(t) = (At+B)e^{3t}$

Explanation: First, find the complementary solution. The characteristic equation is $r^2 - 6r + 9 = 0$, which factors as $(r-3)^2 = 0$. This gives a repeated root $r=3$. The complementary solution is $y_c(t) = c_1e^{3t} + c_2te^{3t}$. The non-homogeneous term is $g(t) = 5e^{3t}$. The initial guess for the particular solution would be $Y_p(t) = Ae^{3t}$. However, this term is present in the complementary solution. According to the modification rule, we must multiply the guess by $t$, giving $Ate^{3t}$. This new guess is also present in the complementary solution. Therefore, we must multiply by $t$ again, which gives the correct form $Y_p(t) = At^2e^{3t}$.

Question 3

1. $Y_p(t) = A + Be^{-3t}$
2. $Y_p(t) = At + Be^{-3t}$
3. $Y_p(t) = A + Bte^{-3t}$
4. $Y_p(t) = At + Bte^{-3t}$ (correct answer)

Explanation: The characteristic equation is $r^2 + 3r = 0$, or $r(r+3)=0$, giving roots $r=0$ and $r=-3$. The complementary solution is $y_c(t) = c_1e^{0t} + c_2e^{-3t} = c_1 + c_2e^{-3t}$. The non-homogeneous term is $g(t)=4 - 2e^{-3t}$. We consider each part. For the constant term 4, the initial guess is $A$. This is a solution to the homogeneous equation (the $c_1$ term), so we must modify it to $At$. For the term $-2e^{-3t}$, the initial guess is $Be^{-3t}$. This is also a solution to the homogeneous equation (the $c_2e^{-3t}$ term), so we must modify it to $Bte^{-3t}$. Combining these gives the full form for the particular solution: $Y_p(t) = At + Bte^{-3t}$.

Question 4

1. $g(t) = 3t^2e^{-2t}$
2. $g(t) = 6e^{-2t}$ (correct answer)
3. $g(t) = (12t^2 - 24t + 6)e^{-2t}$
4. $g(t) = 6te^{-2t}$

Explanation: To find $g(t)$, we must compute $Y_p'' + 4Y_p' + 4Y_p$. Given $Y_p(t) = 3t^2e^{-2t}$. First derivative: $Y_p' = 6te^{-2t} - 6t^2e^{-2t} = (6t-6t^2)e^{-2t}$. Second derivative: $Y_p'' = (6-12t)e^{-2t} - 2(6t-6t^2)e^{-2t} = (6 - 12t - 12t + 12t^2)e^{-2t} = (12t^2 - 24t + 6)e^{-2t}$. Now substitute into the left side of the ODE: $g(t) = (12t^2 - 24t + 6)e^{-2t} + 4(6t-6t^2)e^{-2t} + 4(3t^2e^{-2t})$. Factor out $e^{-2t}$: $g(t) = [ (12t^2 - 24t + 6) + (24t - 24t^2) + (12t^2) ] e^{-2t}$. The terms in the brackets simplify: $(12t^2 - 24t^2 + 12t^2) + (-24t + 24t) + 6 = 6$. Therefore, $g(t) = 6e^{-2t}$.

Question 5

1. $Y_p(t) = A\sin^2(t)$
2. $Y_p(t) = A + B\cos(2t) + C\sin(2t)$
3. $Y_p(t) = A + t(B\cos(2t) + C\sin(2t))$ (correct answer)
4. $Y_p(t) = A\cos(2t) + B\sin(2t)$

Explanation: The method of undetermined coefficients cannot be directly applied to $g(t) = 8\sin^2(t)$. We must first use a trigonometric identity to rewrite $g(t)$ in a suitable form. Using the power-reducing identity $\sin^2(t) = \frac{1 - \cos(2t)}{2}$, the non-homogeneous term becomes $g(t) = 8 \left( \frac{1 - \cos(2t)}{2} \right) = 4 - 4\cos(2t)$. Now we can determine the form of $Y_p(t)$. The characteristic equation is $r^2+4=0$, with roots $r=\pm 2i$, so $y_c(t) = c_1\cos(2t)+c_2\sin(2t)$. For the constant term 4, the guess is $A$. This does not conflict with $y_c$. For the term $-4\cos(2t)$, the guess is $B\cos(2t)+C\sin(2t)$. This form is present in $y_c$, so we must multiply this part of the guess by $t$. The total form for $Y_p(t)$ is the sum of the forms for each part: $Y_p(t) = A + t(B\cos(2t) + C\sin(2t))$.

Question 6

1. $Y_p(t) = -t^3 - 3t^2 - 6t$ (correct answer)
2. $Y_p(t) = t^3 + 3t^2 + 6t$
3. $Y_p(t) = -t^3 - 3t^2$
4. $Y_p(t) = -3t^2 - 6t - 6$

Explanation: The characteristic equation is $r^2 - r = 0$, or $r(r-1)=0$, with roots $r=0, 1$. The complementary solution is $y_c(t) = c_1e^{0t} + c_2e^t = c_1 + c_2e^t$. The non-homogeneous term is $g(t) = 3t^2$. The initial guess for $Y_p$ is a full quadratic, $At^2 + Bt + C$. However, the constant term $C$ is a solution to the homogeneous equation (corresponding to the $c_1$ term). Therefore, we must multiply the entire guess by $t$, giving $Y_p(t) = t(At^2 + Bt + C) = At^3 + Bt^2 + Ct$. We find the derivatives: $Y_p' = 3At^2 + 2Bt + C$ and $Y_p'' = 6At + 2B$. Substituting into the ODE: $(6At + 2B) - (3At^2 + 2Bt + C) = 3t^2$. Grouping terms: $-3At^2 + (6A - 2B)t + (2B - C) = 3t^2$. Equating coefficients gives: $-3A = 3 \implies A = -1$. $6A - 2B = 0 \implies 6(-1) - 2B = 0 \implies B = -3$. $2B - C = 0 \implies 2(-3) - C = 0 \implies C = -6$. Thus, $Y_p(t) = -t^3 - 3t^2 - 6t$.

Question 7

1. $Y_p(t) = At + B + C\cos(2t) + D\sin(2t)$
2. $Y_p(t) = At + B + t(C\cos(2t) + D\sin(2t))$ (correct answer)
3. $Y_p(t) = t(At + B + C\cos(2t) + D\sin(2t))$
4. $Y_p(t) = At + B + Ct\cos(2t) + D\sin(2t)$

Explanation: The characteristic equation for the homogeneous part is $r^2 + 4 = 0$, which has roots $r = \pm 2i$. The complementary solution is $y_c(t) = c_1\cos(2t) + c_2\sin(2t)$. The non-homogeneous term is $g(t) = t + \cos(2t)$. We find the guess for the particular solution by considering each part of $g(t)$ separately. For the term $t$, the guess is $At+B$. For the term $\cos(2t)$, the guess is $C\cos(2t) + D\sin(2t)$. We must check if any part of the guess is a solution to the homogeneous equation. The guess $At+B$ does not overlap with $y_c$. However, the guess $C\cos(2t) + D\sin(2t)$ does overlap with $y_c$. Therefore, this part of the guess must be multiplied by $t$, becoming $t(C\cos(2t) + D\sin(2t))$. The final form for the particular solution is the sum of the individual forms: $Y_p(t) = At + B + t(C\cos(2t) + D\sin(2t))$.

Question 8

1. The guess for $Y_p(t)$ is incorrect and must be multiplied by $t$.
2. The guess for $Y_p(t)$ is correct, and the conclusion is that $A=5t$, which is not a constant.
3. There is a calculation error, as the method of undetermined coefficients cannot fail this way. (correct answer)
4. The equation has no particular solution of the assumed form.

Explanation: The method of undetermined coefficients guarantees a solution of the assumed form exists, provided the form is chosen correctly. Let's verify the calculation. $Y_p = (At+B)e^t$. $Y_p' = Ae^t + (At+B)e^t = (At+A+B)e^t$. $Y_p'' = Ae^t + (At+A+B)e^t = (At+2A+B)e^t$. Substitute into the ODE: $(At+2A+B)e^t - 2(At+A+B)e^t + 2(At+B)e^t = 5te^t$. Divide by $e^t$: $(At+2A+B) - 2(At+A+B) + 2(At+B) = 5t$. Combine terms: $(A-2A+2A)t + (2A+B-2A-2B+2B) = 5t$. This simplifies to $At + B = 5t$. Equating coefficients gives $A=5$ and $B=0$. The calculation in the stem, which leads to $Ae^t=5te^t$, is incorrect. A calculation error must have occurred. The guess does not need modification as $y_c=e^t(c_1\cos t + c_2\sin t)$ has no overlap.

Question 9

1. The reasoning is correct; the entire guess must be multiplied by $t$ to give $Y_p=t(At+B)e^{-t}$. (correct answer)
2. The reasoning is partially correct; only the conflicting term should be multiplied by $t$, giving $Y_p=Ate^{-t} + Bte^{-t}$.
3. The reasoning is incorrect; the term $Ate^{-t}$ does not duplicate a term in $y_c$, so no modification is needed.
4. The reasoning is correct, but the proposed solution is wrong; the modification rule does not apply to polynomial factors.

Explanation: The student's reasoning is correct. The initial guess for a non-homogeneous term of the form $P_n(t)e^{\alpha t}$ is $(A_n t^n + \dots + A_0)e^{\alpha t}$. In this case, $g(t)=te^{-t}$, so the initial guess is $Y_p = (At+B)e^{-t}$. The complementary solution is $y_c = c_1 e^t + c_2 e^{-t}$. The term $Be^{-t}$ in the expansion of the guess, $Ate^{-t}+Be^{-t}$, is a solution to the homogeneous equation. The modification rule states that if any term in the initial guess is a solution to the homogeneous equation, the entire initial guess must be multiplied by $t^k$ until no term is a solution. Therefore, the entire expression $(At+B)e^{-t}$ must be multiplied by $t$. The correct form is $Y_p = t(At+B)e^{-t} = (At^2+Bt)e^{-t}$.

Question 10

1. $Y_p(t) = e^{-t}(A\cos(2t) + B\sin(2t))$
2. $Y_p(t) = Ate^{-t}\cos(2t)$
3. $Y_p(t) = te^{-t}(A\cos(2t) + B\sin(2t))$ (correct answer)
4. $Y_p(t) = A e^{-t} \cos(2t)$

Explanation: The characteristic equation is $r^2 + 2r + 5 = 0$. Using the quadratic formula, the roots are $r = \frac{-2 \pm \sqrt{4-20}}{2} = -1 \pm 2i$. The complementary solution is $y_c(t) = e^{-t}(c_1\cos(2t) + c_2\sin(2t))$. The non-homogeneous term is $g(t) = 3e^{-t}\cos(2t)$. The initial guess for the particular solution, based on the form of $g(t)$, is $Y_p(t) = e^{-t}(A\cos(2t) + B\sin(2t))$. This form is identical to the form of the complementary solution $y_c(t)$. Therefore, the modification rule applies, and the entire initial guess must be multiplied by $t$. The correct form is $Y_p(t) = te^{-t}(A\cos(2t) + B\sin(2t))$.

Question 11

1. $g(t) = 7e^{2t}\cos(t)$ (correct answer)
2. $g(t) = t\cos(t)$
3. $g(t) = 4e^{2t}$
4. $g(t) = t e^{2t} \sin(t)$

Explanation: The form of the particular solution $Y_p(t)$ has been multiplied by $t$. This indicates that the initial, unmodified guess, which would be $e^{2t}(A\cos(t) + B\sin(t))$, is a solution to the homogeneous equation. The given complementary solution $y_c(t)$ confirms this. The method of undetermined coefficients dictates that the initial guess is based on the form of the non-homogeneous term $g(t)$. A term like $g(t) = 7e^{2t}\cos(t)$ (or any linear combination of $e^{2t}\cos(t)$ and $e^{2t}\sin(t)$) would lead to this initial guess, thus requiring multiplication by $t$.

Question 12

1. $Y_p(t) = 2te^t + 1$ (correct answer)
2. $Y_p(t) = 2te^t - 2$
3. $Y_p(t) = 6te^t + 1$
4. $Y_p(t) = te^t + 1$

Explanation: The problem can be solved using the principle of superposition. We are given that for $g_1(t)=3e^t$, the particular solution is $Y_{p1}(t)=te^t$. By linearity, for $g(t)=2g_1(t)=6e^t$, the particular solution will be $2Y_{p1}(t)=2te^t$. Next, we need to find the particular solution for the term $g_2(t)=-2$. The characteristic equation is $r^2+r-2=0$, with roots $r=1, -2$, so $y_c=c_1e^t+c_2e^{-2t}$. For $g_2(t)=-2$, our guess is $Y_{p2}(t)=A$. Since a constant is not part of $y_c$, no modification is needed. Substituting $Y_{p2}=A$ into the ODE gives $0+0-2A = -2$, so $A=1$. By superposition, the total particular solution is $Y_p(t) = 2te^t + 1$.

Question 13

1. The polynomial degree should be increased by one to account for the forcing term structure
2. A cosine and sine term should be included since complex roots are present in the characteristic equation
3. The exponential factor should be $e^{2x}$ instead of $e^x$ to match the characteristic roots
4. The entire expression should be multiplied by $x$ due to resonance with the homogeneous solution (correct answer)

Explanation: When solving non-homogeneous linear differential equations using the method of undetermined coefficients, you must check whether your proposed particular solution overlaps with the homogeneous solution. This overlap creates a critical issue called resonance. First, let's find the homogeneous solution. The characteristic equation $r^2 - 2r + 1 = 0$ factors as $(r-1)^2 = 0$, giving a repeated root $r = 1$. This means the homogeneous solution is $y_h = (C_1 + C_2x)e^x$, which contains terms of the form $e^x$ and $xe^x$. The proposed particular solution $y_p = e^x(Ax^2 + Bx + C)$ contains the term $Ce^x$, which directly overlaps with the homogeneous solution. When you substitute this into the differential equation, the overlapping terms will vanish, making it impossible to solve for the coefficients. To fix this resonance problem, you must multiply the entire proposed solution by $x$ raised to the power equal to the multiplicity of the repeated root. Since we have a repeated root of multiplicity 2, we multiply by $x^2$, giving $y_p = x^2e^x(Ax^2 + Bx + C)$. Choice A is incorrect because the polynomial degree is already appropriate for the forcing term $3x^2 + 4x$. Choice B is wrong since the characteristic equation has real, not complex roots. Choice C is incorrect because the exponential factor $e^x$ correctly matches the forcing term's exponential part. Key strategy: Always solve the homogeneous equation first and check for resonance before finalizing your particular solution form.

Question 14

1. The coefficients $A$, $B$, and $C$ can be uniquely determined since no resonance occurs
2. Both trigonometric and exponential terms create contradictions requiring complete reformulation of the particular solution
3. The trigonometric terms work correctly, but the exponential term creates a contradiction requiring modification
4. The exponential term works correctly, but the trigonometric terms create a contradiction requiring modification (correct answer)

Explanation: When solving non-homogeneous differential equations using the method of undetermined coefficients, you must check whether any terms in your proposed particular solution are already solutions to the homogeneous equation. This is the resonance condition that determines whether your initial guess will work. Let's examine what happens when we substitute $y_p = A\sin(2x) + B\cos(2x) + Ce^{2x}$ into $y'' + 4y = \sin(2x) + 3\cos(2x) + 5e^{2x}$. First, check the homogeneous equation $y'' + 4y = 0$. Its characteristic equation is $r^2 + 4 = 0$, giving $r = \pm 2i$. The homogeneous solutions are $\sin(2x)$ and $\cos(2x)$. For the exponential term $Ce^{2x}$: Since $e^{2x}$ is not a solution to the homogeneous equation ($r = 2$ is not a root), this term works perfectly and $C$ can be determined uniquely. For the trigonometric terms $A\sin(2x) + B\cos(2x)$: These are solutions to the homogeneous equation, creating resonance. When you substitute these into the differential equation, they satisfy $y'' + 4y = 0$, not the non-zero right-hand side. This creates a contradiction that cannot be resolved. Answer D correctly identifies this situation. Answer A misses the resonance problem entirely. Answer B incorrectly claims both terms fail. Answer C reverses which terms work. Study tip: Always solve the homogeneous equation first and compare its solutions to your particular solution guess. If there's overlap, multiply the problematic terms by $x$ (or higher powers) until resonance is eliminated.

Question 15

1. The form is correct and will yield a unique solution for all coefficients
2. The $x\cos(x)$ and $x\sin(x)$ terms are unnecessary since no resonance occurs for these terms
3. The $A\sin(x)$ and $B\cos(x)$ terms will lead to contradictory equations that cannot be satisfied (correct answer)
4. The $C x\cos(x)$ term is sufficient, but the $D x\sin(x)$ term creates an overdetermined system

Explanation: The characteristic equation is $r^2 + 1 = 0$ with roots $r = \pm i$, so $y_h = c_1\cos(x) + c_2\sin(x)$. The forcing term $\sin(x)$ resonates with $y_h$, requiring $x\sin(x)$ and $x\cos(x)$ terms. For $x\cos(x)$, we need $x^2\cos(x)$ and $x^2\sin(x)$ terms due to resonance. The proposed form mixes resonant and non-resonant terms incorrectly. When $A\sin(x) + B\cos(x)$ is substituted, these terms cancel with parts of the left side, leading to $0 = \sin(x)$, which is impossible. Choice A ignores this contradiction. Choice B misidentifies which terms cause resonance. Choice D incorrectly describes the mathematical issue.

Question 16

1. $y_p = Ae^{2x} + Bx + C$
2. $y_p = Ax^2e^{2x} + Bx + C$ (correct answer)
3. $y_p = Axe^{2x} + Bx^2 + Cx$
4. $y_p = Ae^{2x} + Bx^2 + Cx$

Explanation: The homogeneous equation has characteristic equation $r^2 - 4r + 4 = 0$, giving $(r-2)^2 = 0$, so $r = 2$ is a repeated root. The homogeneous solution is $y_h = (c_1 + c_2x)e^{2x}$. For the forcing term $8e^{2x}$, since $e^{2x}$ appears in the homogeneous solution with multiplicity 2, we multiply by $x^2$ to get $Ax^2e^{2x}$. For $12x$, we use $Bx + C$. Choice A fails to account for the repeated root. Choice C incorrectly uses $Bx^2 + Cx$ for the polynomial term. Choice D uses $Bx^2 + Cx$ unnecessarily for the polynomial and doesn't multiply $e^{2x}$ by $x^2$.

Question 17

1. 5 coefficients (correct answer)
2. 4 coefficients
3. 6 coefficients
4. 7 coefficients

Explanation: When solving nonhomogeneous linear differential equations using undetermined coefficients, you need to carefully analyze each term on the right-hand side and account for any overlap with the homogeneous solution. First, find the homogeneous solution. The characteristic equation $r^2 + 4r + 4 = 0$ factors as $(r + 2)^2 = 0$, giving a repeated root $r = -2$. So the homogeneous solution is $y_h = (c_1 + c_2t)e^{-2t}$. Now examine each term in $te^{-2t} + e^{-2t}\cos(3t)$: For $te^{-2t}$: Normally you'd try $Ate^{-2t}$, but since $e^{-2t}$ and $te^{-2t}$ are already in the homogeneous solution, you must multiply by $t^2$ (the multiplicity of the repeated root). This gives $At^3e^{-2t} + Bt^2e^{-2t}$, requiring 2 coefficients. For $e^{-2t}\cos(3t)$: You'd normally try $(C\cos(3t) + D\sin(3t))e^{-2t}$, but since $e^{-2t}$ appears in the homogeneous solution, multiply by $t^2$: $t^2(E\cos(3t) + F\sin(3t))e^{-2t}$, requiring 2 more coefficients. Wait—you also need the $t^2e^{-2t}\sin(3t)$ term that appears when differentiating, giving you one additional coefficient: $G$. Total: 5 coefficients. Answer A (5 coefficients) is correct. Answer B (4 coefficients) misses the adjustment for repeated roots. Answer C (6 coefficients) likely double-counts somewhere. Answer D (7 coefficients) overcomplicates the overlap resolution. Remember: always check if your trial solution overlaps with the homogeneous solution—multiply by the appropriate power of $t$ to eliminate that overlap.