What is the solution for y^{''}-5y^'=25t ?

The solution for y^{''}-5y^'=25t is y=c_{1}e^{5t}+c_{2}-(5t^2)/2-t

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y^{''}-5y^'=25t

y^{^{''}} - 5 y^{^{'}} = 25 t

y = c_{1} e^{5 t} + c_{2} - \frac{5 t ^{2}}{2} - t

Solution steps

y^{''} (t) - 5 y^{'} (t) = 25 t

Solve linear ODE:

y = c_{1} e^{5 t} + c_{2} - \frac{5 t ^{2}}{2} - t

y = c_{1} e^{5 t} + c_{2} - \frac{5 t ^{2}}{2} - t

What is the solution for y^{''}-5y^'=25t ?
The solution for y^{''}-5y^'=25t is y=c_{1}e^{5t}+c_{2}-(5t^2)/2-t