What is the solution for y^{''}-4y^'+4y=0,y(0)=1,y^'(0)=1 ?

The solution for y^{''}-4y^'+4y=0,y(0)=1,y^'(0)=1 is y=e^{2t}-e^{2t}t

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y^{''}-4y^'+4y=0,y(0)=1,y^'(0)=1

y^{^{''}} - 4 y^{^{'}} + 4 y = 0, y (0) = 1, y^{^{'}} (0) = 1

y = e^{2 t} - e^{2 t} t

Solution steps

y^{''} (t) - 4 y^{'} (t) + 4 y = 0

Solve linear ODE:

y = e^{2 t} - e^{2 t} t

y = e^{2 t} - e^{2 t} t

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\int_{0}^{1} (x^{2} + 8) (e^{- x}) d x

x \to 2 - lim (\frac{x - 3}{x - 2})

\int \frac{e ^{x}}{e ^{x} - 1} d x

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

What is the solution for y^{''}-4y^'+4y=0,y(0)=1,y^'(0)=1 ?
The solution for y^{''}-4y^'+4y=0,y(0)=1,y^'(0)=1 is y=e^{2t}-e^{2t}t