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

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

4 y^{^{''}} - 4 y^{^{'}} + y = 16 e^{\frac{1}{2} t}

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{1}{2} t}

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}

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