What is the solution for x^2y^'+2xy=sin(4x) ?

The solution for x^2y^'+2xy=sin(4x) is y=-(cos(4x))/(4x^2)+(c_{1})/(x^2)

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x^2y^'+2xy=sin(4x)

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

y = - \frac{cos ( 4 x )}{4 x ^{2}} + \frac{c _{1}}{x ^{2}}

Solution steps

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

Solve linear ODE:

y = - \frac{cos ( 4 x )}{4 x ^{2}} + \frac{c _{1}}{x ^{2}}

y = - \frac{cos ( 4 x )}{4 x ^{2}} + \frac{c _{1}}{x ^{2}}

\int - 2 x 9 - x^{2} d x

\int \frac{1 + 4 p}{2 - 2 p - 4 p ^{2}} d p

\int_{0}^{4} (\frac{1}{4} x + (x + 1)) d x

\frac{d}{d x} (e^{- \frac{a}{x}})

x \to 3 lim (\frac{4 x - 12}{x - 3})

What is the solution for x^2y^'+2xy=sin(4x) ?
The solution for x^2y^'+2xy=sin(4x) is y=-(cos(4x))/(4x^2)+(c_{1})/(x^2)