What is the solution for 4y^{''}-4y^'+y=16e^{(t/2)} ?

The solution for 4y^{''}-4y^'+y=16e^{(t/2)} is y=c_{1}e^{t/2}+c_{2}e^{t/2}t+2e^{t/2}t^2

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4y^{''}-4y^'+y=16e^{(t/2)}

4 y^{^{''}} - 4 y^{^{'}} + y = 16 e^{(\frac{t}{2})}

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

Solution steps

4 y^{''} (t) - 4 y^{'} (t) + y = 16 e^{(\frac{t}{2})}

Solve linear ODE:

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

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

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What is the solution for 4y^{''}-4y^'+y=16e^{(t/2)} ?
The solution for 4y^{''}-4y^'+y=16e^{(t/2)} is y=c_{1}e^{t/2}+c_{2}e^{t/2}t+2e^{t/2}t^2