What is the solution for y^{''}+9x=-sin(3x) ?

The solution for y^{''}+9x=-sin(3x) is y= 1/9 sin(3x)-(3x^3)/2+c_{1}x+c_{2}

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y^{''}+9x=-sin(3x)

y^{^{''}} + 9 x = - sin (3 x)

y = \frac{1}{9} sin (3 x) - \frac{3 x ^{3}}{2} + c_{1} x + c_{2}

Solution steps

y^{''} (x) + 9 x = - sin (3 x)

Isolate

y^{''} (x) : y^{''} (x) = - sin (3 x) - 9 x

Solve

y^{''} (x) = - sin (3 x) - 9 x : y^{'} (x) = \frac{1}{3} cos (3 x) - \frac{9 x ^{2}}{2} + c_{1}

Solve

y^{'} (x) = \frac{1}{3} cos (3 x) - \frac{9 x ^{2}}{2} + c_{1} : y = \frac{1}{9} sin (3 x) - \frac{3 x ^{3}}{2} + c_{1} x + c_{2}

y = \frac{1}{9} sin (3 x) - \frac{3 x ^{3}}{2} + c_{1} x + c_{2}

\frac{d}{d x} (\frac{1}{16} x^{4} + \frac{1}{2 x ^{2}})

x \to 3 lim (2 x + 3)

\frac{d}{d x} (r (x) \cdot cos (x))

n = 0 \sum \infty (- \frac{1}{2})^{n - 1}

\frac{d}{d x} (cos (\frac{1}{x ^{2}}))

What is the solution for y^{''}+9x=-sin(3x) ?
The solution for y^{''}+9x=-sin(3x) is y= 1/9 sin(3x)-(3x^3)/2+c_{1}x+c_{2}