What is the solution for y^{''}-4y=t^2+1 ?

The solution for y^{''}-4y=t^2+1 is y=c_{1}e^{2t}+c_{2}e^{-2t}-(t^2)/4-3/8

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y^{''}-4y=t^2+1

y^{^{''}} - 4 y = t^{2} + 1

y = c_{1} e^{2 t} + c_{2} e^{- 2 t} - \frac{t ^{2}}{4} - \frac{3}{8}

Solution steps

y^{''} (t) - 4 y = t^{2} + 1

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} e^{- 2 t} - \frac{t ^{2}}{4} - \frac{3}{8}

y = c_{1} e^{2 t} + c_{2} e^{- 2 t} - \frac{t ^{2}}{4} - \frac{3}{8}

What is the solution for y^{''}-4y=t^2+1 ?
The solution for y^{''}-4y=t^2+1 is y=c_{1}e^{2t}+c_{2}e^{-2t}-(t^2)/4-3/8