What is the solution for 4y^{''}-4y^'+y=40e^{t/2} ?

The solution for 4y^{''}-4y^'+y=40e^{t/2} is y=c_{1}e^{t/2}+c_{2}e^{t/2}t+5e^{t/2}t^2

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4y^{''}-4y^'+y=40e^{t/2}

4 y^{^{''}} - 4 y^{^{'}} + y = 40 e^{\frac{t}{2}}

y = c_{1} e^{\frac{t}{2}} + c_{2} e^{\frac{t}{2}} t + 5 e^{\frac{t}{2}} t^{2}

Solution steps

4 y^{''} (t) - 4 y^{'} (t) + y = 40 e^{\frac{t}{2}}

Solve linear ODE:

y = c_{1} e^{\frac{t}{2}} + c_{2} e^{\frac{t}{2}} t + 5 e^{\frac{t}{2}} t^{2}

y = c_{1} e^{\frac{t}{2}} + c_{2} e^{\frac{t}{2}} t + 5 e^{\frac{t}{2}} t^{2}

\frac{d ^{2} y}{d x ^{2}} + \frac{d y}{d x} = sec (x)

y^{^{''}} - y^{^{'}} + 225 y = 15 sin (15 t)

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

y^{^{''}} + 4 y^{^{'}} + 4 y = 6 sin (x)

y^{^{'''}} - y^{^{'}} = sin^{2} (x)

What is the solution for 4y^{''}-4y^'+y=40e^{t/2} ?
The solution for 4y^{''}-4y^'+y=40e^{t/2} is y=c_{1}e^{t/2}+c_{2}e^{t/2}t+5e^{t/2}t^2