What is the solution for 2y^{''}+3y^'-2y=25e^{-2t} ?

The solution for 2y^{''}+3y^'-2y=25e^{-2t} is y=c_{1}e^{t/2}+c_{2}e^{-2t}-5te^{-2t}

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2y^{''}+3y^'-2y=25e^{-2t}

2 y^{^{''}} + 3 y^{^{'}} - 2 y = 25 e^{- 2 t}

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

Solution steps

2 y^{''} (t) + 3 y^{'} (t) - 2 y = 25 e^{- 2 t}

Solve linear ODE:

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

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

\int \frac{5}{9 + 4 r ^{2}} d r

\frac{d}{d x} (e^{k x})

\int_{- 5}^{5} (25 - x^{2}) d x

\int \frac{1}{( 4 x + 5 ) ^{10}} d x

y^{^{''}} + 2 y^{^{'}} + 10 y = 0

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