What is the solution for y^{''}-4y^'-5y=2x+1,y(0)=1,y^'(0)=2 ?

The solution for y^{''}-4y^'-5y=2x+1,y(0)=1,y^'(0)=2 is y=(-1/3+22/25)e^{5x}+1/3 e^{-x}-(2x)/5+3/25

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y^{''}-4y^'-5y=2x+1,y(0)=1,y^'(0)=2

y^{^{''}} - 4 y^{^{'}} - 5 y = 2 x + 1, y (0) = 1, y^{^{'}} (0) = 2

y = (- \frac{1}{3} + \frac{22}{25}) e^{5 x} + \frac{1}{3} e^{- x} - \frac{2 x}{5} + \frac{3}{25}

Solution steps

y^{''} (x) - 4 y^{'} (x) - 5 y = 2 x + 1

Solve linear ODE:

y = (- \frac{1}{3} + \frac{22}{25}) e^{5 x} + \frac{1}{3} e^{- x} - \frac{2 x}{5} + \frac{3}{25}

y = (- \frac{1}{3} + \frac{22}{25}) e^{5 x} + \frac{1}{3} e^{- x} - \frac{2 x}{5} + \frac{3}{25}

What is the solution for y^{''}-4y^'-5y=2x+1,y(0)=1,y^'(0)=2 ?
The solution for y^{''}-4y^'-5y=2x+1,y(0)=1,y^'(0)=2 is y=(-1/3+22/25)e^{5x}+1/3 e^{-x}-(2x)/5+3/25