What is the solution for y^{''}-3y^'-4y=0,y(0)=a,y^'(0)=-20 ?

The solution for y^{''}-3y^'-4y=0,y(0)=a,y^'(0)=-20 is y=e^{4t}(a+(-20-4a)/5)-(e^{-t}(-20-4a))/5

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y^{''}-3y^'-4y=0,y(0)=a,y^'(0)=-20

y^{^{''}} - 3 y^{^{'}} - 4 y = 0, y (0) = a, y^{^{'}} (0) = - 20

y = e^{4 t} (a + \frac{- 20 - 4 a}{5}) - \frac{e ^{- t} ( - 20 - 4 a )}{5}

Solution steps

y^{''} (t) - 3 y^{'} (t) - 4 y = 0

Solve linear ODE:

y = e^{4 t} (a + \frac{- 20 - 4 a}{5}) - \frac{e ^{- t} ( - 20 - 4 a )}{5}

y = e^{4 t} (a + \frac{- 20 - 4 a}{5}) - \frac{e ^{- t} ( - 20 - 4 a )}{5}

\frac{d}{d x} (x^{\frac{8}{3}} - 20 x^{\frac{2}{3}})

\frac{d y}{d x} = y^{2} sin (x^{2}), y (- 2) = \frac{1}{3}

f (r) = π r^{2}

2 x x, at x = 1

6 x

What is the solution for y^{''}-3y^'-4y=0,y(0)=a,y^'(0)=-20 ?
The solution for y^{''}-3y^'-4y=0,y(0)=a,y^'(0)=-20 is y=e^{4t}(a+(-20-4a)/5)-(e^{-t}(-20-4a))/5