A Second-Order Initial-Value Problem with Complex Roots

Question

Solve $y'' - 4y' + 13y = 0$ with $y(0) = 2$ and $y'(0) = 1$ .

a. $y = e^{2x}\left(2\cos 3x + \tfrac{1}{3}\sin 3x\right)$
b. $y = e^{2x}\left(2\cos 3x - \sin 3x\right)$
c. $y = \tfrac{1}{2}e^{5x} + \tfrac{3}{2}e^{-x}$
d. $y = e^{3x}\left(2\cos 2x - \tfrac{5}{2}\sin 2x\right)$

Correct answer

B. $y = e^{2x}\left(2\cos 3x - \sin 3x\right)$

Full reasoning

1The equation has constant coefficients, so substitute $y = e^{rx}$ to get the characteristic equation $r^{2} - 4r + 13 = 0$ .
2Solve it: $r = \dfrac{4 \pm \sqrt{16 - 52}}{2} = \dfrac{4 \pm \sqrt{-36}}{2} = \dfrac{4 \pm 6i}{2} = 2 \pm 3i$ , so $\alpha = 2$ and $\beta = 3$ .
3The complex-root case gives the general solution $y = e^{2x}\left(A\cos 3x + B\sin 3x\right)$ .
4Apply $y(0) = 2$ : at $x = 0$ the expression is $1 \cdot (A \cdot 1 + B \cdot 0) = A$ , so $A = 2$ .
5Differentiate with the product rule: $y' = e^{2x}\left[\left(2A + 3B\right)\cos 3x + \left(2B - 3A\right)\sin 3x\right]$ , so $y'(0) = 2A + 3B$ .
6Apply $y'(0) = 1$ : $2(2) + 3B = 1$ gives $3B = -3$ , so $B = -1$ and $y = e^{2x}\left(2\cos 3x - \sin 3x\right)$ .

Why each choice is right or wrong

Choice A

This differentiates $e^{2x}\left(A\cos 3x + B\sin 3x\right)$ without the product rule, using $y'(0) = 3B$ instead of $y'(0) = 2A + 3B$ . The exponential factor contributes $2A$ to $y'(0)$ and cannot be ignored.

Choice B

Correct. The roots are $2 \pm 3i$ , so $y = e^{2x}\left(A\cos 3x + B\sin 3x\right)$ ; $y(0) = A = 2$ and $y'(0) = 2A + 3B = 1$ give $B = -1$ .

Choice C

This reads $\sqrt{-36}$ as $6$ instead of $6i$ , producing the fictitious real roots $r = 2 \pm 3$ . The discriminant $16 - 52 = -36$ is negative, so the roots are complex and the solution oscillates.

Choice D

This swaps the real and imaginary parts of the root, using $3 \pm 2i$ instead of $2 \pm 3i$ . The real part is the exponential growth rate and the imaginary part is the oscillation frequency, so the two are not interchangeable.

Related formula

Complex-root case of the characteristic equation

r = \alpha \pm \beta i \;\Longrightarrow\; y = e^{\alpha x}\left(c_1 \cos \beta x + c_2 \sin \beta x\right)

Assumptions: The equation is homogeneous with constant coefficients and the discriminant is negative. Differentiating this form requires the product rule, so y'(0) = alpha*c1 + beta*c2.

Sources

Authored for this public explainer registry. It has no database question ID and is not copied from a protected or official exam bank.

Calculus Volume 3, Section 7.1: Second-Order Linear Equations — OpenStax. Accessed 2026-08-15. OpenStax textbook content is CC BY-NC-SA 4.0; attribute and avoid verbatim reuse beyond short cited references.

Share this worked solution

Sources and corrections

Sources last checked 2026-08-15

Every source cited on this page was checked on the date shown, and we update the page when a source changes. If something looks wrong, tell us and we'll recheck it.

Report an issue with this page