What is the solution for xy^'+2y^'-(x+2)sin(x)+y=0 ?

The solution for xy^'+2y^'-(x+2)sin(x)+y=0 is y=-cos(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^'-(x+2)sin(x)+y=0

x y^{^{'}} + 2 y^{^{'}} - (x + 2) sin (x) + y = 0

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

Solution steps

x y^{'} (x) + 2 y^{'} (x) - (x + 2) sin (x) + y = 0

Solve linear ODE:

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

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

\frac{\partial}{\partial x} ((x^{2} y - y^{3})^{5})

- \frac{e ^{(- 2 s)}}{( s ^{2} - 4 s + 4 )}

\int x^{5} e^{4 x} d x

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

\frac{d}{d x} (tan (x) + x sec^{2} (x))

What is the solution for xy^'+2y^'-(x+2)sin(x)+y=0 ?
The solution for xy^'+2y^'-(x+2)sin(x)+y=0 is y=-cos(x)+(sin(x))/(x+2)+(c_{1})/(x+2)