What is the solution for y^{''}-5y^'+4y=-3t^2 ?

The solution for y^{''}-5y^'+4y=-3t^2 is y=c_{1}e^{4t}+c_{2}e^t-(3t^2)/4-(15t)/8-63/32

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y^{''}-5y^'+4y=-3t^2

y^{^{''}} - 5 y^{^{'}} + 4 y = - 3 t^{2}

y = c_{1} e^{4 t} + c_{2} e^{t} - \frac{3 t ^{2}}{4} - \frac{15 t}{8} - \frac{63}{32}

Solution steps

y^{''} (t) - 5 y^{'} (t) + 4 y = - 3 t^{2}

Solve linear ODE:

y = c_{1} e^{4 t} + c_{2} e^{t} - \frac{3 t ^{2}}{4} - \frac{15 t}{8} - \frac{63}{32}

y = c_{1} e^{4 t} + c_{2} e^{t} - \frac{3 t ^{2}}{4} - \frac{15 t}{8} - \frac{63}{32}

What is the solution for y^{''}-5y^'+4y=-3t^2 ?
The solution for y^{''}-5y^'+4y=-3t^2 is y=c_{1}e^{4t}+c_{2}e^t-(3t^2)/4-(15t)/8-63/32