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

The solution for y^{''}-2y^'+5y=1+x,y(0)=0,y^'(0)=4 is y=e^x(-7/25 cos(2x)+51/25 sin(2x))+x/5+7/25

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

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

y = e^{x} (- \frac{7}{25} cos (2 x) + \frac{51}{25} sin (2 x)) + \frac{x}{5} + \frac{7}{25}

Solution steps

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

Solve linear ODE:

y = e^{x} (- \frac{7}{25} cos (2 x) + \frac{51}{25} sin (2 x)) + \frac{x}{5} + \frac{7}{25}

y = e^{x} (- \frac{7}{25} cos (2 x) + \frac{51}{25} sin (2 x)) + \frac{x}{5} + \frac{7}{25}

What is the solution for y^{''}-2y^'+5y=1+x,y(0)=0,y^'(0)=4 ?
The solution for y^{''}-2y^'+5y=1+x,y(0)=0,y^'(0)=4 is y=e^x(-7/25 cos(2x)+51/25 sin(2x))+x/5+7/25