What is the solution for x^2y^'+2xy=17cos^2(x) ?

The solution for x^2y^'+2xy=17cos^2(x) is y=(17(2x+sin(2x)))/(4x^2)+(c_{1})/(x^2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

x^2y^'+2xy=17cos^2(x)

x^{2} y^{^{'}} + 2 x y = 17 cos^{2} (x)

y = \frac{17 ( 2 x + sin ( 2 x ) )}{4 x ^{2}} + \frac{c _{1}}{x ^{2}}

Solution steps

x^{2} y^{'} (x) + 2 x y = 17 cos^{2} (x)

Solve linear ODE:

y = \frac{17 ( 2 x + sin ( 2 x ) )}{4 x ^{2}} + \frac{c _{1}}{x ^{2}}

y = \frac{17 ( 2 x + sin ( 2 x ) )}{4 x ^{2}} + \frac{c _{1}}{x ^{2}}

(x - 9) e^{x}

\int \frac{sin ^{3} ( x )}{1 - cos ^{2} ( x )} d x

\int u^{\frac{1}{3}} d u

\frac{\partial}{\partial x} (sin (x) - cos (x y))

\frac{\partial}{\partial x} (63 x^{3} cos (9 x^{2} y))

What is the solution for x^2y^'+2xy=17cos^2(x) ?
The solution for x^2y^'+2xy=17cos^2(x) is y=(17(2x+sin(2x)))/(4x^2)+(c_{1})/(x^2)