What is the solution for 2y^{''}+3y^'=-1 ?

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

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2y^{''}+3y^'=-1

2 y^{^{''}} + 3 y^{^{'}} = - 1

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

Solution steps

2 y^{''} (t) + 3 y^{'} (t) = - 1

Solve linear ODE:

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

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

y^{^{''}} + 5 y^{^{'}} + 6 y = x e^{x}

y^{^{''}} - 10 y^{^{'}} + 21 y = 205 x^{2} - 100 x + 10

- x^{^{''}} + x^{^{'}} = 6 e^{t} + 2 t - 2

y^{^{''}} + 5 y^{^{'}} + 6 y = 12 e^{- 5 t} + 12 e^{- t} + 24

\frac{d ^{2} y}{d x ^{2}} - y = cos (2 x)

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