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}

16 y^{^{''''}} - y = e^{\frac{x}{2}}

y^{^{''}} + 5 y^{^{'}} + 6 y = 36 e^{t}, y (0) = - 5, y^{^{'}} (0) = 21

y^{^{''}} - 5 y^{^{'}} + 6 y = 2 x + 3 e^{8 x}

y^{^{''}} - 6 y^{^{'}} + 8 y = 2 x - 1

y^{^{''}} + 2 y^{^{'}} - 3 y = x^{2} e^{x}

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