What is the solution for 9y^{''}+24y^'+16y=8e^x ?

The solution for 9y^{''}+24y^'+16y=8e^x is y=c_{1}e^{-(4x)/3}+c_{2}e^{-(4x)/3}x+8/49 e^x

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9y^{''}+24y^'+16y=8e^x

9 y^{^{''}} + 24 y^{^{'}} + 16 y = 8 e^{x}

y = c_{1} e^{- \frac{4 x}{3}} + c_{2} e^{- \frac{4 x}{3}} x + \frac{8}{49} e^{x}

Solution steps

9 y^{''} (x) + 24 y^{'} (x) + 16 y = 8 e^{x}

Solve linear ODE:

y = c_{1} e^{- \frac{4 x}{3}} + c_{2} e^{- \frac{4 x}{3}} x + \frac{8}{49} e^{x}

y = c_{1} e^{- \frac{4 x}{3}} + c_{2} e^{- \frac{4 x}{3}} x + \frac{8}{49} e^{x}

y^{^{''}} - 4 y^{^{'}} + 4 y = 25 cos (x)

x^{^{''}} - x = e^{- t} sin (e^{- t}) + cos (e^{- t})

y^{^{''}} - 2 y^{^{'}} - 3 y = 2 e x - 10

y^{^{'''}} + 3 y^{^{''}} + y^{^{'}} - 5 y = - 5 x

y^{^{''}} + 4 y = 4 sin (2 t) + 8 cos (2 t)

What is the solution for 9y^{''}+24y^'+16y=8e^x ?
The solution for 9y^{''}+24y^'+16y=8e^x is y=c_{1}e^{-(4x)/3}+c_{2}e^{-(4x)/3}x+8/49 e^x