What is the solution for y^{''}-y^'+1/4 y=xe^{x/2} ?

The solution for y^{''}-y^'+1/4 y=xe^{x/2} is y=c_{1}e^{x/2}+c_{2}e^{x/2}x+(e^{x/2}x^3)/6

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y^{''}-y^'+1/4 y=xe^{x/2}

y^{^{''}} - y^{^{'}} + \frac{1}{4} y = x e^{\frac{x}{2}}

y = c_{1} e^{\frac{x}{2}} + c_{2} e^{\frac{x}{2}} x + \frac{e ^{\frac{x}{2}} x ^{3}}{6}

Solution steps

y^{''} (x) - y^{'} (x) + \frac{1}{4} y = x e^{\frac{x}{2}}

Solve linear ODE:

y = c_{1} e^{\frac{x}{2}} + c_{2} e^{\frac{x}{2}} x + \frac{e ^{\frac{x}{2}} x ^{3}}{6}

y = c_{1} e^{\frac{x}{2}} + c_{2} e^{\frac{x}{2}} x + \frac{e ^{\frac{x}{2}} x ^{3}}{6}

What is the solution for y^{''}-y^'+1/4 y=xe^{x/2} ?
The solution for y^{''}-y^'+1/4 y=xe^{x/2} is y=c_{1}e^{x/2}+c_{2}e^{x/2}x+(e^{x/2}x^3)/6