What is the solution for y^{''}-4y^'+4y=1 ?

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

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y^{''}-4y^'+4y=1

y^{^{''}} - 4 y^{^{'}} + 4 y = 1

y = c_{1} e^{2 t} + c_{2} t e^{2 t} + \frac{1}{4}

Solution steps

y^{''} (t) - 4 y^{'} (t) + 4 y = 1

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} t e^{2 t} + \frac{1}{4}

y = c_{1} e^{2 t} + c_{2} t e^{2 t} + \frac{1}{4}

\int_{0}^{\infty} e^{- s t} d t

y = (\frac{( 3 x ^{2} )}{2 x + 4})^{3}

x d y - y d x = (x y)^{\frac{1}{2}} d x

f (x) = 7 cos (x), at (\frac{2 π}{3})

x \to 1 lim (\frac{2 - x}{x - 1})

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