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

The solution for y^{''}-3y^'=4t is y=c_{1}e^{3t}+c_{2}-(2t^2)/3-(4t)/9

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y^{''}-3y^'=4t

y^{^{''}} - 3 y^{^{'}} = 4 t

y = c_{1} e^{3 t} + c_{2} - \frac{2 t ^{2}}{3} - \frac{4 t}{9}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{3 t} + c_{2} - \frac{2 t ^{2}}{3} - \frac{4 t}{9}

y = c_{1} e^{3 t} + c_{2} - \frac{2 t ^{2}}{3} - \frac{4 t}{9}

\int 2 e^{7} d x

\int_{- 3}^{- 2} \frac{x + 2}{x ^{2} + 4 x} d x

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y^{^{''}} - y^{^{'}} - 2 y = 3 e^{2 x}

\int_{- 2}^{2} (3 - (x^{2} - 1)) d x

What is the solution for y^{''}-3y^'=4t ?
The solution for y^{''}-3y^'=4t is y=c_{1}e^{3t}+c_{2}-(2t^2)/3-(4t)/9