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

The solution for y^{''}+25y=5x,y(0)=1,y^'(0)=5 is y=cos(5x)+24/25 sin(5x)+x/5

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

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

y = cos (5 x) + \frac{24}{25} sin (5 x) + \frac{x}{5}

Solution steps

y^{''} (x) + 25 y = 5 x

Solve linear ODE:

y = cos (5 x) + \frac{24}{25} sin (5 x) + \frac{x}{5}

y = cos (5 x) + \frac{24}{25} sin (5 x) + \frac{x}{5}

y^{^{''}} - 2 y^{^{'}} - 3 y = 8 e x - 3

y^{^{''}} + 2 y^{^{'}} + 4 y = 6 cos (3 x)

x^{^{''}} + 4 x^{^{'}} + 3 x = 1, x^{^{'}} (0) = 0

y^{^{'''}} + y^{^{'}} = sec (t), y^{^{'''}} (0) = - 2

\frac{d ^{2} y}{d x ^{2}} + y = 2 e^{3 x}

What is the solution for y^{''}+25y=5x,y(0)=1,y^'(0)=5 ?
The solution for y^{''}+25y=5x,y(0)=1,y^'(0)=5 is y=cos(5x)+24/25 sin(5x)+x/5