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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\int arcsin (\frac{2 x}{5}) d x

\int \frac{1}{1 - x ^{4}} d x

7 e^{u}

\frac{d}{d x} (7^{3^{x^{2}}})

4 + 7 cos (5 t)

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