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

The solution for y^{''}-y^'+1/4 y=2+e^{x/2} is y=c_{1}e^{x/2}+c_{2}e^{x/2}x+8+(e^{x/2}x^2)/2

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\int \frac{3 - x}{x ^{3} + 4 x ^{2} + 3 x} d x

\frac{d}{d x} (4 ln (cos (x)))

x \to 3 - lim (2 x - 1)

\int_{1}^{4} 5 t ln (t) d t

x \to - 3 lim (- \frac{5}{( x + 3 ) ^{2}})

What is the solution for y^{''}-y^'+1/4 y=2+e^{x/2} ?
The solution for y^{''}-y^'+1/4 y=2+e^{x/2} is y=c_{1}e^{x/2}+c_{2}e^{x/2}x+8+(e^{x/2}x^2)/2