What is the solution for y^{''}+6y^'+8y=e^{5x},y(0)=2 ?

The solution for y^{''}+6y^'+8y=e^{5x},y(0)=2 is y=(-c_{2}+125/63)e^{-2x}+c_{2}e^{-4x}+1/63 e^{5x}

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y^{''}+6y^'+8y=e^{5x},y(0)=2

y^{^{''}} + 6 y^{^{'}} + 8 y = e^{5 x}, y (0) = 2

y = (- c_{2} + \frac{125}{63}) e^{- 2 x} + c_{2} e^{- 4 x} + \frac{1}{63} e^{5 x}

Solution steps

y^{''} (x) + 6 y^{'} (x) + 8 y = e^{5 x}

Solve linear ODE:

y = (- c_{2} + \frac{125}{63}) e^{- 2 x} + c_{2} e^{- 4 x} + \frac{1}{63} e^{5 x}

y = (- c_{2} + \frac{125}{63}) e^{- 2 x} + c_{2} e^{- 4 x} + \frac{1}{63} e^{5 x}

\frac{d y}{d x} + 7 y = 12

\frac{sin ( x )}{1 - sin ( x )}

\frac{( 5 x + 2 ) ^{5}}{( 4 x + 1 ) ^{4}}

y^{^{''}} + 4 y = 5

\int \frac{1}{3 x - 4} d x

What is the solution for y^{''}+6y^'+8y=e^{5x},y(0)=2 ?
The solution for y^{''}+6y^'+8y=e^{5x},y(0)=2 is y=(-c_{2}+125/63)e^{-2x}+c_{2}e^{-4x}+1/63 e^{5x}