Question

5. The general solution of $y'' - 2y' + y = 2e^{3x} + 8e^{-3x}$ is: (a) $y = C_1e^x + C_2xe^x + cosh 3x$ (b) $y = C_1e^x + C_2xe^x + 2e^{3x} - 2e^{-3x}$ (c) $y = C_1e^x + C_2xe^x + frac{1}{4}e^{3x} - frac{1}{2}e^{-3x}$ (d) $y = C_1e^x + C_2xe^x + frac{1}{2}e^{3x} + frac{1}{4}e^{-3x}$ (e) None of the above.

          5. The general solution of $y'' - 2y' + y = 2e^{3x} + 8e^{-3x}$ is:
(a) $y = C_1e^x + C_2xe^x + cosh 3x$
(b) $y = C_1e^x + C_2xe^x + 2e^{3x} - 2e^{-3x}$
(c) $y = C_1e^x + C_2xe^x + frac{1}{4}e^{3x} - frac{1}{2}e^{-3x}$
(d) $y = C_1e^x + C_2xe^x + frac{1}{2}e^{3x} + frac{1}{4}e^{-3x}$
(e) None of the above.

$5. The general solution of y” - 2y' + y = 2e^3x + 8e^-3x is: (a) y = C1e^x + C2xe^x + cosh 3x (b) y = C1e^x + C2xe^x + 2e^3x - 2e^-3x (c) y = C1e^x + C2xe^x + frac14e^3x - frac12e^-3x (d) y = C1e^x + C2xe^x + frac12e^3x + frac14e^-3x (e) None of the above.$

Added by Ramon D.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Shaiju T

03/22/2022

Step 1

Step 1: Write down the differential equation given: $y'' + 2y' + y = 2e^{3x} + 8e^{-3x}$. Show more…

Show all steps

Thanks for your feedback!

The general solution of y'' + 2y' + y = 2e^(3x) + 8e^(-3x) is: (a) y = C1e^x + C2e^(-x) + cosh(3x) (b) y = C1e^(x) + C2e^(-x) + 2e^(3x) + 1 (c) y = C1e^(x) + C2e^(-x) + e^(3x) + 2e^(3x) (d) y = C1e^(x) + C2e^(-x) + 2e^(3x) + 1e^(-3x) (e) None of the above.