What is the solution for x^2y^'+2xy=15cos^2(x) ?

The solution for x^2y^'+2xy=15cos^2(x) is y=(15(2x+sin(2x)))/(4x^2)+(c_{1})/(x^2)

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x^2y^'+2xy=15cos^2(x)

x^{2} y^{^{'}} + 2 x y = 15 cos^{2} (x)

y = \frac{15 ( 2 x + sin ( 2 x ) )}{4 x ^{2}} + \frac{c _{1}}{x ^{2}}

Solution steps

x^{2} y^{'} (x) + 2 x y = 15 cos^{2} (x)

Solve linear ODE:

y = \frac{15 ( 2 x + sin ( 2 x ) )}{4 x ^{2}} + \frac{c _{1}}{x ^{2}}

y = \frac{15 ( 2 x + sin ( 2 x ) )}{4 x ^{2}} + \frac{c _{1}}{x ^{2}}

\int_{- 1}^{2} x^{2} - 2 x d x

f (x) = 4 - x

x \to 0 + lim (x^{3 s i n (x)})

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

\int \frac{x ^{2}}{( x - 1 ) ^{2}} d x

What is the solution for x^2y^'+2xy=15cos^2(x) ?
The solution for x^2y^'+2xy=15cos^2(x) is y=(15(2x+sin(2x)))/(4x^2)+(c_{1})/(x^2)