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

The solution for y^{''}+4y^'-3y=x is y=c_{1}e^{(-2+sqrt(7))x}+c_{2}e^{(-2-sqrt(7))x}-x/3-4/9

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y^{''}+4y^'-3y=x

y^{^{''}} + 4 y^{^{'}} - 3 y = x

y = c_{1} e^{(- 2 + 7) x} + c_{2} e^{(- 2 - 7) x} - \frac{x}{3} - \frac{4}{9}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{(- 2 + 7) x} + c_{2} e^{(- 2 - 7) x} - \frac{x}{3} - \frac{4}{9}

y = c_{1} e^{(- 2 + 7) x} + c_{2} e^{(- 2 - 7) x} - \frac{x}{3} - \frac{4}{9}

\frac{d}{d x} (sin^{2} (x) + cos^{2} (x))

\int e^{- \frac{t}{2}} d t

y^{^{''}} = 8 t^{2}

\frac{d}{d x} (\frac{x}{x ^{3} + 2})

n = 0 \sum \infty n e^{- \frac{n}{2}}

What is the solution for y^{''}+4y^'-3y=x ?
The solution for y^{''}+4y^'-3y=x is y=c_{1}e^{(-2+sqrt(7))x}+c_{2}e^{(-2-sqrt(7))x}-x/3-4/9