What is the solution for y^{''}-2y^'+y=15e^x ?

The solution for y^{''}-2y^'+y=15e^x is y=c_{1}e^x+c_{2}xe^x+(15e^xx^2)/2

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y^{''}-2y^'+y=15e^x

y^{^{''}} - 2 y^{^{'}} + y = 15 e^{x}

y = c_{1} e^{x} + c_{2} x e^{x} + \frac{15 e ^{x} x ^{2}}{2}

Solution steps

y^{''} (x) - 2 y^{'} (x) + y = 15 e^{x}

Solve linear ODE:

y = c_{1} e^{x} + c_{2} x e^{x} + \frac{15 e ^{x} x ^{2}}{2}

y = c_{1} e^{x} + c_{2} x e^{x} + \frac{15 e ^{x} x ^{2}}{2}

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

y^{^{''}} + y^{^{'}} = 1, y (0) = 1, y^{^{'}} (0) = 1

D (x) = 72 - 1.8 x

y^{^{'''}} - 2 y^{^{''}} + y^{^{'}} = (3 x - 4) e^{- 2 x} - 5 e^{x}

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

What is the solution for y^{''}-2y^'+y=15e^x ?
The solution for y^{''}-2y^'+y=15e^x is y=c_{1}e^x+c_{2}xe^x+(15e^xx^2)/2