What is the solution for y^{''}+3y^'-10y=-42t+20t^2 ?

The solution for y^{''}+3y^'-10y=-42t+20t^2 is y=c_{1}e^{2t}+c_{2}e^{-5t}-2t^2+3t+1/2

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y^{''}+3y^'-10y=-42t+20t^2

y^{^{''}} + 3 y^{^{'}} - 10 y = - 42 t + 20 t^{2}

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

Solution steps

y^{''} (t) + 3 y^{'} (t) - 10 y = - 42 t + 20 t^{2}

Solve linear ODE:

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

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

y^{^{''}} - 4 y^{^{'}} + 4 y = 2 e^{5 t}

4 q^{^{''}} + 8 q^{^{'}} + 80 q = 88

y^{^{''}} + y = sin (e^{- x}) (e^{- 2 x})

\frac{d ^{2} y}{d x ^{2}} - 25 y = 3 x^{2} - 2 + e^{x}

\frac{d ^{2} y}{d x ^{2}} - 4 (\frac{d y}{d x}) + 8 y = x e^{x}

What is the solution for y^{''}+3y^'-10y=-42t+20t^2 ?
The solution for y^{''}+3y^'-10y=-42t+20t^2 is y=c_{1}e^{2t}+c_{2}e^{-5t}-2t^2+3t+1/2