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

The solution for y^{''}+4y^'=t^2+3e^t is y=c_{1}+c_{2}e^{-4t}+(t^3)/(12)-(t^2)/(16)+t/(32)+3/5 e^t

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y^{''}+4y^'=t^2+3e^t

y^{^{''}} + 4 y^{^{'}} = t^{2} + 3 e^{t}

y = c_{1} + c_{2} e^{- 4 t} + \frac{t ^{3}}{12} - \frac{t ^{2}}{16} + \frac{t}{32} + \frac{3}{5} e^{t}

Solution steps

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

Solve linear ODE:

y = c_{1} + c_{2} e^{- 4 t} + \frac{t ^{3}}{12} - \frac{t ^{2}}{16} + \frac{t}{32} + \frac{3}{5} e^{t}

y = c_{1} + c_{2} e^{- 4 t} + \frac{t ^{3}}{12} - \frac{t ^{2}}{16} + \frac{t}{32} + \frac{3}{5} e^{t}

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x \to 3 lim (\frac{1 - x}{x ^{2} - 9})

d y - (y - 9)^{2} \cdot d x = 0

\frac{d}{d x} (\frac{cos ( m x )}{sin ( n x )})

What is the solution for y^{''}+4y^'=t^2+3e^t ?
The solution for y^{''}+4y^'=t^2+3e^t is y=c_{1}+c_{2}e^{-4t}+(t^3)/(12)-(t^2)/(16)+t/(32)+3/5 e^t