Question

Please solve the following differential equation:​

Answer 1

Solve the corresponding homogeneous equation:

$y'' = 0$

Integrate twice to get the characteristic solution:

$\displaystyle \int \frac{d^2y}{dt^2} \, dt = \int 0 \, dt \implies \dfrac{dy}{dt} = C_1$

$\displaystyle \int \frac{dy}{dt} \, dt = \int C_1 \, dt \implies y = C_1 t + C_2$

For the particular solution, consider the ansatz

$y = a e^{-2t} + b e^{4t}$

with second derivative

$y'' = 4a e^{-2t} + 16 be^{4t}$

Substitute this into the differential equation and solve for the unknown coefficients.

$4a e^{-2t} + 16 be^{4t} = e^{-2t} + 10e^{4t}$

$\implies \begin{cases}4a = 1 \\ 16b = 10\end{cases} \implies a = \dfrac14, b = \dfrac58$

The general solution is then

$y = C_1 t + C_2 + \dfrac14 e^{-2t} + \dfrac58 e^{4t}$

with first derivative

$y' = C_1 - \dfrac12 e^{-2t} + \dfrac52 e^{4t}$

Use the initial conditions to solve for the remaining constants.

$y(0) = 1 \implies 1 = C_2 + \dfrac14 + \dfrac58 \implies C_2 = \dfrac18$

$y'(0) = 0 \implies 0 = C_1 - \dfrac12 + \dfrac52 \implies C_1 = -2$

Then the particular solution to the initial value problem is

$\boxed{y = -2t + \dfrac18 + \dfrac14 e^{-2t} + \dfrac58 e^{4t}}$

(A)