What is the solution for x^2y^'+2xy=sin(3x) ?

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

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x^2y^'+2xy=sin(3x)

x^{2} y^{^{'}} + 2 x y = sin (3 x)

y = - \frac{cos ( 3 x )}{3 x ^{2}} + \frac{c _{1}}{x ^{2}}

Solution steps

x^{2} y^{'} (x) + 2 x y = sin (3 x)

Solve linear ODE:

y = - \frac{cos ( 3 x )}{3 x ^{2}} + \frac{c _{1}}{x ^{2}}

y = - \frac{cos ( 3 x )}{3 x ^{2}} + \frac{c _{1}}{x ^{2}}

x \to 0 - lim (\frac{x + 94}{x})

\frac{d}{d t} (\frac{2 t}{t ^{2} + 1})

e^{5 x} cos (8 x)

h \to 0 lim ((2 h)^{2})

\frac{d}{d x} (sin^{c o s (x)} (x))

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