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

Popular >

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}}

\int \frac{48}{( 6 x + 1 ) ^{9}} d x

\int 10 sin^{3} (x co) s^{7} x d x

\frac{4 x}{x}

\int \frac{x}{2 x - 6} d x

4 x y^{^{'}} - 20 y = x^{- 8}

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)