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}

\frac{4}{2 - 4} + \frac{27}{4 + 2}

\int 7 e^{3 x + e^{3 x}} d x

\frac{\partial}{\partial y} (\frac{( x + y )}{( x - y )})

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\int \frac{1}{2 u + 1} d u

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