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

The solution for y^{''}-10y^'+25y=2e^5x is y=c_{1}e^{5x}+c_{2}xe^{5x}+(2e^5x)/(25)+(4e^5)/(125)

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y^{''}-10y^'+25y=2e^5x

y^{^{''}} - 10 y^{^{'}} + 25 y = 2 e^{5} x

y = c_{1} e^{5 x} + c_{2} x e^{5 x} + \frac{2 e ^{5} x}{25} + \frac{4 e ^{5}}{125}

Solution steps

y^{''} (x) - 10 y^{'} (x) + 25 y = 2 e^{5} x

Solve linear ODE:

y = c_{1} e^{5 x} + c_{2} x e^{5 x} + \frac{2 e ^{5} x}{25} + \frac{4 e ^{5}}{125}

y = c_{1} e^{5 x} + c_{2} x e^{5 x} + \frac{2 e ^{5} x}{25} + \frac{4 e ^{5}}{125}

6 y^{^{''}} + y^{^{'}} - y = - 25 e^{2 x} + 7 e^{- 2 x}

(d^{3} - 3 d - 2) y = x^{2}

y^{^{''}} - 3 y^{^{'}} - 4 = - 8 e^{t} cos (2 t)

y^{^{''}} - 64 y = - 192 t + 128 t^{2}

y^{^{'''}} - 5 y^{^{''}} + 6 y^{^{'}} = 2 x + 3 e^{2 x}

What is the solution for y^{''}-10y^'+25y=2e^5x ?
The solution for y^{''}-10y^'+25y=2e^5x is y=c_{1}e^{5x}+c_{2}xe^{5x}+(2e^5x)/(25)+(4e^5)/(125)