Solve the corresponding homogeneous equation:
[tex]y'' = 0[/tex]
Integrate twice to get the characteristic solution:
[tex]\displaystyle \int \frac{d^2y}{dt^2} \, dt = \int 0 \, dt \implies \dfrac{dy}{dt} = C_1[/tex]
[tex]\displaystyle \int \frac{dy}{dt} \, dt = \int C_1 \, dt \implies y = C_1 t + C_2[/tex]
For the particular solution, consider the ansatz
[tex]y = a e^{-2t} + b e^{4t}[/tex]
with second derivative
[tex]y'' = 4a e^{-2t} + 16 be^{4t}[/tex]
Substitute this into the differential equation and solve for the unknown coefficients.
[tex]4a e^{-2t} + 16 be^{4t} = e^{-2t} + 10e^{4t}[/tex]
[tex]\implies \begin{cases}4a = 1 \\ 16b = 10\end{cases} \implies a = \dfrac14, b = \dfrac58[/tex]
The general solution is then
[tex]y = C_1 t + C_2 + \dfrac14 e^{-2t} + \dfrac58 e^{4t}[/tex]
with first derivative
[tex]y' = C_1 - \dfrac12 e^{-2t} + \dfrac52 e^{4t}[/tex]
Use the initial conditions to solve for the remaining constants.
[tex]y(0) = 1 \implies 1 = C_2 + \dfrac14 + \dfrac58 \implies C_2 = \dfrac18[/tex]
[tex]y'(0) = 0 \implies 0 = C_1 - \dfrac12 + \dfrac52 \implies C_1 = -2[/tex]
Then the particular solution to the initial value problem is
[tex]\boxed{y = -2t + \dfrac18 + \dfrac14 e^{-2t} + \dfrac58 e^{4t}}[/tex]
(A)
