What is the solution for y^{''}-4y^'-3y=8e^x ?

The solution for y^{''}-4y^'-3y=8e^x is y=c_{1}e^{(2+sqrt(7))x}+c_{2}e^{(2-sqrt(7))x}-4/3 e^x

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y^{''}-4y^'-3y=8e^x

y^{^{''}} - 4 y^{^{'}} - 3 y = 8 e^{x}

y = c_{1} e^{(2 + 7) x} + c_{2} e^{(2 - 7) x} - \frac{4}{3} e^{x}

Solution steps

y^{''} (x) - 4 y^{'} (x) - 3 y = 8 e^{x}

Solve linear ODE:

y = c_{1} e^{(2 + 7) x} + c_{2} e^{(2 - 7) x} - \frac{4}{3} e^{x}

y = c_{1} e^{(2 + 7) x} + c_{2} e^{(2 - 7) x} - \frac{4}{3} e^{x}

y^{^{''}} + 6 y^{^{'}} + y = 2 sin (3 t)

y^{^{''}} - 9 y^{^{'}} = 54

y^{^{''}} + y^{^{'}} - 2 y = 1 + x

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

y^{^{''}} + 6 y^{^{'}} + 9 y = e^{3 x} ln (x)

What is the solution for y^{''}-4y^'-3y=8e^x ?
The solution for y^{''}-4y^'-3y=8e^x is y=c_{1}e^{(2+sqrt(7))x}+c_{2}e^{(2-sqrt(7))x}-4/3 e^x