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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

2 x^{2} + x y + 2 y^{2} = 5, (1, 1)

\int x (x^{2} + 4)^{2} d x

\int \frac{sec ^{4} ( t )}{tan ^{2} ( t )} d t

\int \frac{tan ( ln ( 5 x + 6 ) )}{5 x + 6} d x

\frac{d}{d x} (\frac{x ^{2} + 2 x}{4 - 5 x})

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