What is the solution for y^{''}+9y=7x,y(0)=1,y^'(0)=3 ?

The solution for y^{''}+9y=7x,y(0)=1,y^'(0)=3 is y=cos(3x)+20/27 sin(3x)+(7x)/9

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y^{''}+9y=7x,y(0)=1,y^'(0)=3

y^{^{''}} + 9 y = 7 x, y (0) = 1, y^{^{'}} (0) = 3

y = cos (3 x) + \frac{20}{27} sin (3 x) + \frac{7 x}{9}

Solution steps

y^{''} (x) + 9 y = 7 x

Solve linear ODE:

y = cos (3 x) + \frac{20}{27} sin (3 x) + \frac{7 x}{9}

y = cos (3 x) + \frac{20}{27} sin (3 x) + \frac{7 x}{9}

y^{^{''}} - 0.01 y = - 0.01 \cdot 20

y^{^{''}} + \frac{1}{2} y^{^{'}} + y = 4 sin (\frac{x}{2})

y^{^{'''}} - y = x^{3} - 1

8 y^{^{''}} + y = t, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} - 4 y = t e^{4 t}

What is the solution for y^{''}+9y=7x,y(0)=1,y^'(0)=3 ?
The solution for y^{''}+9y=7x,y(0)=1,y^'(0)=3 is y=cos(3x)+20/27 sin(3x)+(7x)/9