What is the solution for x^2y^'+4xy=(sin(2x))/(x^2) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

x^2y^'+4xy=(sin(2x))/(x^2)

x^{2} y^{^{'}} + 4 x y = \frac{sin ( 2 x )}{x ^{2}}

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

Solution steps

x^{2} y^{'} (x) + 4 x y = \frac{sin ( 2 x )}{x ^{2}}

Solve linear ODE:

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

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

\frac{d}{d x} (C e^{- x})

\int \frac{w}{( 1 + w ) ^{\frac{3}{4}}} d w

y^{^{'}} = 6 x^{2} - 3 x^{2} y

\frac{d}{d y} (x^{- 2} y^{- 5})

x \to 0 lim (\frac{4}{sin ( 5 x )})

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