What is the solution for (dy}{dx}=\frac{(y-sqrt(x^2+y^2)))/x ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(dy}{dx}=\frac{(y-sqrt(x^2+y^2)))/x

\frac{d y}{d x} = \frac{( y - x ^{2} + y ^{2} )}{x}

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

Solution steps

\frac{d y}{d x} = \frac{( y - x ^{2} + y ^{2} )}{x}

Solve using substitution:

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

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

y y^{^{'}} = (y^{2} - y) (2 x + 5)

\int sin (3 x) - cos (3 x) d x

y^{^{''}} + 4 y^{^{'}} + 5 y = 10

\frac{d}{d x} (3 x^{2} + 2 e^{x})

h (x) = sec (- 4 x^{2} - 3 x + 2)

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