What is the solution for y^{''}-10y^'+25y=4e^{5x} ?

The solution for y^{''}-10y^'+25y=4e^{5x} is y=c_{1}e^{5x}+c_{2}xe^{5x}+2x^2e^{5x}

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y^{''}-10y^'+25y=4e^{5x}

y^{^{''}} - 10 y^{^{'}} + 25 y = 4 e^{5 x}

y = c_{1} e^{5 x} + c_{2} x e^{5 x} + 2 x^{2} e^{5 x}

Solution steps

y^{''} (x) - 10 y^{'} (x) + 25 y = 4 e^{5 x}

Solve linear ODE:

y = c_{1} e^{5 x} + c_{2} x e^{5 x} + 2 x^{2} e^{5 x}

y = c_{1} e^{5 x} + c_{2} x e^{5 x} + 2 x^{2} e^{5 x}

y^{^{''}} + 10 y^{^{'}} + 25 y = 50 sin (5 x)

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

y^{^{''}} + 6 y^{^{'}} + 34 y = 30 cos (2 t)

y^{^{''}} - 7 y^{^{'}} + 9 y = x e^{x}

y^{^{''''}} - 16 y = x e^{2 x}

What is the solution for y^{''}-10y^'+25y=4e^{5x} ?
The solution for y^{''}-10y^'+25y=4e^{5x} is y=c_{1}e^{5x}+c_{2}xe^{5x}+2x^2e^{5x}