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}}

\frac{d y}{d x} = sec^{2} (x) (2 + y)^{2}

\frac{4 e ^{- 2 s}}{3 s ^{2} + 48}

(2 y^{2} x - 3) d x + (2 y x^{2} + 4) d y = 0

\frac{s + 6}{s ^{2} + 6 s + 25}

\int \frac{t - 3}{t + 1} d 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)