What is the solution for 4y^{''}-4y^'+y=48e^{t/2} ?

The solution for 4y^{''}-4y^'+y=48e^{t/2} is y=c_{1}e^{t/2}+c_{2}e^{t/2}t+6e^{t/2}t^2

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4y^{''}-4y^'+y=48e^{t/2}

4 y^{^{''}} - 4 y^{^{'}} + y = 48 e^{\frac{t}{2}}

y = c_{1} e^{\frac{t}{2}} + c_{2} e^{\frac{t}{2}} t + 6 e^{\frac{t}{2}} t^{2}

Solution steps

4 y^{''} (t) - 4 y^{'} (t) + y = 48 e^{\frac{t}{2}}

Solve linear ODE:

y = c_{1} e^{\frac{t}{2}} + c_{2} e^{\frac{t}{2}} t + 6 e^{\frac{t}{2}} t^{2}

y = c_{1} e^{\frac{t}{2}} + c_{2} e^{\frac{t}{2}} t + 6 e^{\frac{t}{2}} t^{2}

y^{^{''}} + 3 y^{^{'}} + 5 y = sin (t), y (0) = 1, y^{^{'}} (0) = 0

y^{^{''}} - y = 5 sin (2 t)

y^{^{''}} + y^{^{'}} - 6 y = 3 e^{2 t}

y^{^{''}} - 4 y = 6 x - 4 x^{3}

y^{^{'''}} - y^{^{''}} + 4 y^{^{'}} - 4 y = - 3 e^{t} + 4 e^{2 t}

What is the solution for 4y^{''}-4y^'+y=48e^{t/2} ?
The solution for 4y^{''}-4y^'+y=48e^{t/2} is y=c_{1}e^{t/2}+c_{2}e^{t/2}t+6e^{t/2}t^2