What is the solution for y^{''}-4y^'=16t+36,y(-2)=2,y^'(-2)=-1 ?

The solution for y^{''}-4y^'=16t+36,y(-2)=2,y^'(-2)=-1 is y=(e^8)/4 e^{4t}-41/4-2t^2-10t

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y^{''}-4y^'=16t+36,y(-2)=2,y^'(-2)=-1

y^{^{''}} - 4 y^{^{'}} = 16 t + 36, y (- 2) = 2, y^{^{'}} (- 2) = - 1

y = \frac{e ^{8}}{4} e^{4 t} - \frac{41}{4} - 2 t^{2} - 10 t

Solution steps

y^{''} (t) - 4 y^{'} (t) = 16 t + 36

Solve linear ODE:

y = \frac{e ^{8}}{4} e^{4 t} - \frac{41}{4} - 2 t^{2} - 10 t

y = \frac{e ^{8}}{4} e^{4 t} - \frac{41}{4} - 2 t^{2} - 10 t

\frac{d ^{2} y}{d x ^{2}} + a \frac{d y}{d x} = b

y^{^{''''}} - 5 y^{^{''}} + 4 y = 20 e^{3 x}

y^{^{''}} + 36 y = 12 tan (6 t)

(- 8 x^{^{''}} - 16 x^{^{'}} + 24) = 8 e^{2 t}

y^{^{''}} - y^{^{'}} + 2 y = 4 e^{x}

What is the solution for y^{''}-4y^'=16t+36,y(-2)=2,y^'(-2)=-1 ?
The solution for y^{''}-4y^'=16t+36,y(-2)=2,y^'(-2)=-1 is y=(e^8)/4 e^{4t}-41/4-2t^2-10t