What is the solution for y^{''}+2y^'-35y=x^2 ?

The solution for y^{''}+2y^'-35y=x^2 is y=c_{1}e^{5x}+c_{2}e^{-7x}-(x^2)/(35)-(4x)/(1225)-78/42875

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y^{''}+2y^'-35y=x^2

y^{^{''}} + 2 y^{^{'}} - 35 y = x^{2}

y = c_{1} e^{5 x} + c_{2} e^{- 7 x} - \frac{x ^{2}}{35} - \frac{4 x}{1225} - \frac{78}{42875}

Solution steps

y^{''} (x) + 2 y^{'} (x) - 35 y = x^{2}

Solve linear ODE:

y = c_{1} e^{5 x} + c_{2} e^{- 7 x} - \frac{x ^{2}}{35} - \frac{4 x}{1225} - \frac{78}{42875}

y = c_{1} e^{5 x} + c_{2} e^{- 7 x} - \frac{x ^{2}}{35} - \frac{4 x}{1225} - \frac{78}{42875}

y^{^{''}} + 16 y = cos (3 t), y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} - y^{^{'}} - 6 y = - 14 t + 12 t^{2}

4 y^{^{''}} - 4 y^{^{'}} + y = 16 e \frac{t}{2}

(D^{2} - 2 D + 1) y = x e^{x} sin (x)

y^{^{''}} + 100 y = 200 sin (10 x), y^{^{'}} (0) = 0, y^{^{''}} (0) = 0

What is the solution for y^{''}+2y^'-35y=x^2 ?
The solution for y^{''}+2y^'-35y=x^2 is y=c_{1}e^{5x}+c_{2}e^{-7x}-(x^2)/(35)-(4x)/(1225)-78/42875