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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

d er i v a t i v e f (x) = (9 x^{2} - 15 x) e^{x}

\int sin (x) cos (x) ln (sin (x)) d x

\int_{0}^{1} 4 - x^{2} d x

\int \frac{9}{x ^{4}} d x

a re a 7 x - x^{2}, 2 x

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