What is the solution for y^{''}-7y^'+10y=2xe^x ?

The solution for y^{''}-7y^'+10y=2xe^x is y=c_{1}e^{5x}+c_{2}e^{2x}+(e^xx)/2+5/8 e^x

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y^{''}-7y^'+10y=2xe^x

y^{^{''}} - 7 y^{^{'}} + 10 y = 2 x e^{x}

y = c_{1} e^{5 x} + c_{2} e^{2 x} + \frac{e ^{x} x}{2} + \frac{5}{8} e^{x}

Solution steps

y^{''} (x) - 7 y^{'} (x) + 10 y = 2 x e^{x}

Solve linear ODE:

y = c_{1} e^{5 x} + c_{2} e^{2 x} + \frac{e ^{x} x}{2} + \frac{5}{8} e^{x}

y = c_{1} e^{5 x} + c_{2} e^{2 x} + \frac{e ^{x} x}{2} + \frac{5}{8} e^{x}

What is the solution for y^{''}-7y^'+10y=2xe^x ?
The solution for y^{''}-7y^'+10y=2xe^x is y=c_{1}e^{5x}+c_{2}e^{2x}+(e^xx)/2+5/8 e^x