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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

s l o p e h (x) = 4^{x} - 6

d er i v a t i v e g (x) = 3 x^{2} cos (x) + 8 x

\int \frac{x}{( x ^{2} + 9 )} d x

l a pl a ce t r an s f or m x - 5

\int \frac{g}{\frac{1}{p} - \frac{g z}{k}} d z

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