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

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

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4y^{''}-4y^'+y=8e^{x/2}

4 y^{^{''}} - 4 y^{^{'}} + y = 8 e^{\frac{x}{2}}

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

Solution steps

4 y^{''} (x) - 4 y^{'} (x) + y = 8 e^{\frac{x}{2}}

Solve linear ODE:

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

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

y = 8 e^{17 t}

4 csc (5 x)

f (t) = 4 t^{2}

\frac{\partial}{\partial x} (A e^{i (k x - wt)})

\frac{\partial}{\partial x} (ln (sin^{2} (x)))

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