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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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