What is the solution for y^{''}+y^'-2y=0,y(0)=-5,y^'(0)=-1 ?

The solution for y^{''}+y^'-2y=0,y(0)=-5,y^'(0)=-1 is y=(-5+4/3)e^t-4/3 e^{-2t}

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y^{''}+y^'-2y=0,y(0)=-5,y^'(0)=-1

y^{^{''}} + y^{^{'}} - 2 y = 0, y (0) = - 5, y^{^{'}} (0) = - 1

y = (- 5 + \frac{4}{3}) e^{t} - \frac{4}{3} e^{- 2 t}

Solution steps

y^{''} (t) + y^{'} (t) - 2 y = 0

Solve linear ODE:

y = (- 5 + \frac{4}{3}) e^{t} - \frac{4}{3} e^{- 2 t}

y = (- 5 + \frac{4}{3}) e^{t} - \frac{4}{3} e^{- 2 t}

\int (2 y sin (x) + e^{2 z}) d y

y^{^{''}} + 12 y^{^{'}} + 45 y = 0

\int_{1}^{2} \frac{9 x ^{3} + 4 x ^{2} - 2}{x} d x

\int \frac{sec ( x )}{tan ^{2} ( x )} d x

n = 1 \sum \infty 7 (3)^{n}

What is the solution for y^{''}+y^'-2y=0,y(0)=-5,y^'(0)=-1 ?
The solution for y^{''}+y^'-2y=0,y(0)=-5,y^'(0)=-1 is y=(-5+4/3)e^t-4/3 e^{-2t}