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

The solution for y^{''}-y^'+1/4 y=4+e^{x/2} is y=c_{1}e^{x/2}+c_{2}e^{x/2}x+16+(e^{x/2}x^2)/2

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

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

y = c_{1} e^{\frac{x}{2}} + c_{2} e^{\frac{x}{2}} x + 16 + \frac{e ^{\frac{x}{2}} x ^{2}}{2}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{\frac{x}{2}} + c_{2} e^{\frac{x}{2}} x + 16 + \frac{e ^{\frac{x}{2}} x ^{2}}{2}

y = c_{1} e^{\frac{x}{2}} + c_{2} e^{\frac{x}{2}} x + 16 + \frac{e ^{\frac{x}{2}} x ^{2}}{2}

n = 1 \sum \infty \frac{( 4 ^{n} + 1 )}{5 ^{n}}

\int - x e^{- 2 x} d x

5 - x^{2}

cos (π x), at \frac{1}{2}

y = ln (3 x + 1) e^{x^{2} + 4}

What is the solution for y^{''}-y^'+1/4 y=4+e^{x/2} ?
The solution for y^{''}-y^'+1/4 y=4+e^{x/2} is y=c_{1}e^{x/2}+c_{2}e^{x/2}x+16+(e^{x/2}x^2)/2