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+8et+32e

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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 + 8 e t + 32 e

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 + 8 e t + 32 e

y = c_{1} e^{\frac{t}{2}} + c_{2} e^{\frac{t}{2}} t + 8 e t + 32 e

(D^{2} - 2 D + 1) y = x e^{x} sin (x)

y^{^{''}} + 100 y = 200 sin (10 x), y^{^{'}} (0) = 0, y^{^{''}} (0) = 0

y^{^{''}} + 9 y = cos (3 x) + sin (3 x), y (0) = 2, y^{^{'}} (0) = 1

y^{^{''}} - 6 y^{^{'}} + 9 y = t^{- 6} e^{3 t}

y^{^{''}} + 5 y^{^{'}} + 6 y = 12 e^{t}, y (0) = - 11, y (0) = 31

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+8et+32e