What is the solution for y^{''}+4y^'+3y=15te^{3t} ?

The solution for y^{''}+4y^'+3y=15te^{3t} is y=c_{1}e^{-t}+c_{2}e^{-3t}+(5e^{3t}t)/8-25/96 e^{3t}

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y^{''}+4y^'+3y=15te^{3t}

y^{^{''}} + 4 y^{^{'}} + 3 y = 15 t e^{3 t}

y = c_{1} e^{- t} + c_{2} e^{- 3 t} + \frac{5 e ^{3 t} t}{8} - \frac{25}{96} e^{3 t}

Solution steps

y^{''} (t) + 4 y^{'} (t) + 3 y = 15 t e^{3 t}

Solve linear ODE:

y = c_{1} e^{- t} + c_{2} e^{- 3 t} + \frac{5 e ^{3 t} t}{8} - \frac{25}{96} e^{3 t}

y = c_{1} e^{- t} + c_{2} e^{- 3 t} + \frac{5 e ^{3 t} t}{8} - \frac{25}{96} e^{3 t}

x^{2} = 4 y, 3 x - 4 y + 4 = 0

y^{^{'}} + y = e^{- t} sin (t), y (0) = 1

y^{^{'}} = y (1 - y)

x \to - 4 lim (8 x^{2} + 1)

- 2 sin (4 x)

What is the solution for y^{''}+4y^'+3y=15te^{3t} ?
The solution for y^{''}+4y^'+3y=15te^{3t} is y=c_{1}e^{-t}+c_{2}e^{-3t}+(5e^{3t}t)/8-25/96 e^{3t}