What is the solution for x^2y^'+2xy=19cos^2(x) ?

The solution for x^2y^'+2xy=19cos^2(x) is y=(19(2x+sin(2x)))/(4x^2)+(c_{1})/(x^2)

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x^2y^'+2xy=19cos^2(x)

x^{2} y^{^{'}} + 2 x y = 19 cos^{2} (x)

y = \frac{19 ( 2 x + sin ( 2 x ) )}{4 x ^{2}} + \frac{c _{1}}{x ^{2}}

Solution steps

x^{2} y^{'} (x) + 2 x y = 19 cos^{2} (x)

Solve linear ODE:

y = \frac{19 ( 2 x + sin ( 2 x ) )}{4 x ^{2}} + \frac{c _{1}}{x ^{2}}

y = \frac{19 ( 2 x + sin ( 2 x ) )}{4 x ^{2}} + \frac{c _{1}}{x ^{2}}

y^{^{''}} + 8 y^{^{'}} - 9 y = 0, y (1) = 1, y^{^{'}} (1) = 0

\int r 16 - r^{2} d r

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\int (1 - cos (x))^{3} d x

What is the solution for x^2y^'+2xy=19cos^2(x) ?
The solution for x^2y^'+2xy=19cos^2(x) is y=(19(2x+sin(2x)))/(4x^2)+(c_{1})/(x^2)