What is the solution for (y+sqrt(x^2+y^2))dx=xdy ?

The solution for (y+sqrt(x^2+y^2))dx=xdy is y=-1/(2e^{c_{1)}}+(e^{c_{1}}x^2)/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(y+sqrt(x^2+y^2))dx=xdy

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

y = - \frac{1}{2 e ^{c_{1}}} + \frac{e ^{c_{1}} x ^{2}}{2}

Solution steps

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

Solve using substitution:

y = - \frac{1}{2 e ^{c_{1}}} + \frac{e ^{c_{1}} x ^{2}}{2}

y = - \frac{1}{2 e ^{c_{1}}} + \frac{e ^{c_{1}} x ^{2}}{2}

\int \frac{6 x ^{2}}{( x ^{6} + 1 ) ( x ^{3} + 1 )} d x

x \to 0 lim (sin (9 x))

f (x) = 2 x^{7}

\frac{d ^{10}}{d x ^{10}} (\frac{x}{e ^{x}})

What is the solution for (y+sqrt(x^2+y^2))dx=xdy ?
The solution for (y+sqrt(x^2+y^2))dx=xdy is y=-1/(2e^{c_{1)}}+(e^{c_{1}}x^2)/2