What is the solution for y^{''}+25y=2x,y(0)=1,y^'(0)=5 ?

The solution for y^{''}+25y=2x,y(0)=1,y^'(0)=5 is y=cos(5x)+123/125 sin(5x)+(2x)/(25)

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y^{''}+25y=2x,y(0)=1,y^'(0)=5

y^{^{''}} + 25 y = 2 x, y (0) = 1, y^{^{'}} (0) = 5

y = cos (5 x) + \frac{123}{125} sin (5 x) + \frac{2 x}{25}

Solution steps

y^{''} (x) + 25 y = 2 x

Solve linear ODE:

y = cos (5 x) + \frac{123}{125} sin (5 x) + \frac{2 x}{25}

y = cos (5 x) + \frac{123}{125} sin (5 x) + \frac{2 x}{25}

What is the solution for y^{''}+25y=2x,y(0)=1,y^'(0)=5 ?
The solution for y^{''}+25y=2x,y(0)=1,y^'(0)=5 is y=cos(5x)+123/125 sin(5x)+(2x)/(25)