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

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

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4y^{''}-4y^'+y=16e^{(x/2)}

4 y^{^{''}} - 4 y^{^{'}} + y = 16 e^{(\frac{x}{2})}

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

Solution steps

4 y^{''} (x) - 4 y^{'} (x) + y = 16 e^{(\frac{x}{2})}

Solve linear ODE:

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

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

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