What is the solution for y^{''}+25y=6x,y(0)=1,y^'(0)=5 ?

The solution for y^{''}+25y=6x,y(0)=1,y^'(0)=5 is y=cos(5x)+119/125 sin(5x)+(6x)/(25)

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y^{''}+25y=6x,y(0)=1,y^'(0)=5

y^{^{''}} + 25 y = 6 x, y (0) = 1, y^{^{'}} (0) = 5

y = cos (5 x) + \frac{119}{125} sin (5 x) + \frac{6 x}{25}

Solution steps

y^{''} (x) + 25 y = 6 x

Solve linear ODE:

y = cos (5 x) + \frac{119}{125} sin (5 x) + \frac{6 x}{25}

y = cos (5 x) + \frac{119}{125} sin (5 x) + \frac{6 x}{25}

y^{^{'''}} - y^{^{''}} + y^{^{'}} - y = e^{t} (t^{2} + sin (t))

y^{^{''}} + 16 y = 3, y (0) = 0, y^{^{'}} (0) = 0

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

y^{^{''}} - 12 y^{^{'}} + 36 y = 42 x + 2

2 y^{^{''}} - 3 y^{^{'}} - 2 y = e^{2 x} + x^{2}

What is the solution for y^{''}+25y=6x,y(0)=1,y^'(0)=5 ?
The solution for y^{''}+25y=6x,y(0)=1,y^'(0)=5 is y=cos(5x)+119/125 sin(5x)+(6x)/(25)