What is the solution for y^{''}-3y^'-4y=0,y(0)=a,y^'(0)=-5 ?

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

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y^{''}-3y^'-4y=0,y(0)=a,y^'(0)=-5

y^{^{''}} - 3 y^{^{'}} - 4 y = 0, y (0) = a, y^{^{'}} (0) = - 5

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

Solution steps

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

Solve linear ODE:

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

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

\int (t^{2} - 6)^{2} d t

f (x) = \frac{x ^{2} + 5 x}{x + 2}

x \to \infty lim (x + 3)

\frac{d}{d x} (x^{3} cos (x y))

\frac{t ^{3}}{a}

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