What is the solution for y^{''}-12y^'+36y=(x+1)e^x ?

The solution for y^{''}-12y^'+36y=(x+1)e^x is y=c_{1}e^{6x}+c_{2}xe^{6x}+(e^xx)/(25)+7/125 e^x

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y^{''}-12y^'+36y=(x+1)e^x

y^{^{''}} - 12 y^{^{'}} + 36 y = (x + 1) e^{x}

y = c_{1} e^{6 x} + c_{2} x e^{6 x} + \frac{e ^{x} x}{25} + \frac{7}{125} e^{x}

Solution steps

y^{''} (x) - 12 y^{'} (x) + 36 y = (x + 1) e^{x}

Solve linear ODE:

y = c_{1} e^{6 x} + c_{2} x e^{6 x} + \frac{e ^{x} x}{25} + \frac{7}{125} e^{x}

y = c_{1} e^{6 x} + c_{2} x e^{6 x} + \frac{e ^{x} x}{25} + \frac{7}{125} e^{x}

What is the solution for y^{''}-12y^'+36y=(x+1)e^x ?
The solution for y^{''}-12y^'+36y=(x+1)e^x is y=c_{1}e^{6x}+c_{2}xe^{6x}+(e^xx)/(25)+7/125 e^x