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} ((1 + 2 sin (x)) cos (x))

n = 1 \sum \infty \frac{n}{( n + 1 ) 3 ^{n}}

\int cos^{2} (x) \cdot sin (2 x) d x

\int \frac{- x - 1}{2 ( x ^{2} + 1 )} d x

\int 20 cot^{4} (x) d 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)