What is the solution for xy^'+y^'-(x+1)cos(x)+y=0 ?

The solution for xy^'+y^'-(x+1)cos(x)+y=0 is y=sin(x)+(cos(x))/(x+1)+(c_{1})/(x+1)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

xy^'+y^'-(x+1)cos(x)+y=0

x y^{^{'}} + y^{^{'}} - (x + 1) cos (x) + y = 0

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

Solution steps

x y^{'} (x) + y^{'} (x) - (x + 1) cos (x) + y = 0

Solve linear ODE:

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

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

\frac{d}{d x} ((e^{- x} + e^{x})^{3})

\int \frac{x}{( x - 3 )} d x

\frac{s ^{2} - 1}{( s ^{2} + 1 ) ^{2}}

\frac{d}{d x} (\frac{( a x + b y ( x ) )}{c x + d y})

\frac{d x}{d t} = 5 x (9 - x)

What is the solution for xy^'+y^'-(x+1)cos(x)+y=0 ?
The solution for xy^'+y^'-(x+1)cos(x)+y=0 is y=sin(x)+(cos(x))/(x+1)+(c_{1})/(x+1)