What is the solution for y^{''}+25y=3x,y(0)=1,y^'(0)=5 ?

The solution for y^{''}+25y=3x,y(0)=1,y^'(0)=5 is y=cos(5x)+122/125 sin(5x)+(3x)/(25)

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y^{''}+25y=3x,y(0)=1,y^'(0)=5

y^{^{''}} + 25 y = 3 x, y (0) = 1, y^{^{'}} (0) = 5

y = cos (5 x) + \frac{122}{125} sin (5 x) + \frac{3 x}{25}

Solution steps

y^{''} (x) + 25 y = 3 x

Solve linear ODE:

y = cos (5 x) + \frac{122}{125} sin (5 x) + \frac{3 x}{25}

y = cos (5 x) + \frac{122}{125} sin (5 x) + \frac{3 x}{25}

y^{^{'''}} + 7 y^{^{''}} + 15 y^{^{'}} + 9 = 1

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

x^{^{''}} + 9 x = 8 sin (3 t)

\frac{d ^{2} y}{d x ^{2}} - 4 y = 10 e^{3 x}

x^{^{''}} + 2 x^{^{'}} + x = cos (2 t) + 3 sin (2 t)

What is the solution for y^{''}+25y=3x,y(0)=1,y^'(0)=5 ?
The solution for y^{''}+25y=3x,y(0)=1,y^'(0)=5 is y=cos(5x)+122/125 sin(5x)+(3x)/(25)