What is the solution for y^{''}-y^'+1/4 y=7+e^{x/2} ?

The solution for y^{''}-y^'+1/4 y=7+e^{x/2} is y=c_{1}e^{x/2}+c_{2}e^{x/2}x+28+(e^{x/2}x^2)/2

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y^{''}-y^'+1/4 y=7+e^{x/2}

y^{^{''}} - y^{^{'}} + \frac{1}{4} y = 7 + e^{\frac{x}{2}}

y = c_{1} e^{\frac{x}{2}} + c_{2} e^{\frac{x}{2}} x + 28 + \frac{e ^{\frac{x}{2}} x ^{2}}{2}

Solution steps

y^{''} (x) - y^{'} (x) + \frac{1}{4} y = 7 + e^{\frac{x}{2}}

Solve linear ODE:

y = c_{1} e^{\frac{x}{2}} + c_{2} e^{\frac{x}{2}} x + 28 + \frac{e ^{\frac{x}{2}} x ^{2}}{2}

y = c_{1} e^{\frac{x}{2}} + c_{2} e^{\frac{x}{2}} x + 28 + \frac{e ^{\frac{x}{2}} x ^{2}}{2}

V^{^{''}} + 25 \cdot 1 0^{2} V^{^{'}} + 25 \cdot 1 0^{6} V = 5 \cdot 1 0^{8}

y^{^{''}} + 49 y = 6 cos (7 x) + 9 sin (7 x)

3 y^{^{''}} - 6 y^{^{'}} + 30 y = 15 sin (x)

y^{^{''}} - 2 y^{^{'}} + 26 y = e^{x} cos (5 x)

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

What is the solution for y^{''}-y^'+1/4 y=7+e^{x/2} ?
The solution for y^{''}-y^'+1/4 y=7+e^{x/2} is y=c_{1}e^{x/2}+c_{2}e^{x/2}x+28+(e^{x/2}x^2)/2