What is the solution for (x^2+1)dy+4xydx=xdx ?

The solution for (x^2+1)dy+4xydx=xdx is y=(c_{1})/(4(x^4+2x^2+1))+1/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(x^2+1)dy+4xydx=xdx

(x^{2} + 1) d y + 4 x y d x = x d x

y = \frac{c _{1}}{4 ( x ^{4} + 2 x ^{2} + 1 )} + \frac{1}{4}

Solution steps

(x^{2} + 1) d y + 4 x y d x = x d x

Solve separable ODE:

y = \frac{c _{1}}{4 ( x ^{4} + 2 x ^{2} + 1 )} + \frac{1}{4}

y = \frac{c _{1}}{4 ( x ^{4} + 2 x ^{2} + 1 )} + \frac{1}{4}

\int \frac{- 4 x}{( 1 - 2 x ^{2} ) ^{3}} d x

f (x) = 2 tan (x), at x = \frac{π}{6}

\frac{d}{d x} (e^{\frac{1}{2 x}})

\int tan (b) x d x

f (x) = tan (e^{x})

What is the solution for (x^2+1)dy+4xydx=xdx ?
The solution for (x^2+1)dy+4xydx=xdx is y=(c_{1})/(4(x^4+2x^2+1))+1/4