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+2e^{t/2}t^2

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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 + 2 e^{\frac{t}{2}} t^{2}

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 + 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}

f (x) = 3 e^{x} + 2 x, (0, 3)

x \to 0 lim (\frac{1}{x ^{3}})

x \to 0 + lim (x^{17 x})

\int_{0}^{5} x e^{- x^{2}} d x

\int x cos (118 x) d x

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+2e^{t/2}t^2