Hint: Let x=yu, Then dxdy=ydudy+u ∴ydudy+u+yuy2u2+y2=y ydudy+u+uu2+1=y ydudy=y−u−uu2+1 (y−u−uu2+1)dydu=y This belongs to an Abel equation of the second kind. Let v=y−u−uu2+1 Then y=v+u+uu2+1 dydu=dvdu+1+1(u2+1)32 ∴v(dvdu+1+1(u2+1)32)=v+u+uu2+1 vdvdu+(1+1(u2+1)32)v=v+u+uu2+1 vdvdu=−v(u2+1)32+u+uu2+1

Step by step solution to "Solve dxdy+xx2+y2=y"

Addison Gross

Answered question

2022-01-28

Solve

\frac{d x}{d y} + \frac{x}{\sqrt{x^{2} + y^{2}}} = y

Answer & Explanation

helsinka04

Beginner2022-01-29Added 11 answers

Hint:
Let x=yu,
Then

\frac{d x}{d y} = y \frac{d u}{d y} + u

∴ y \frac{d u}{d y} + u + \frac{y u}{\sqrt{y^{2} u^{2} + y^{2}}} = y

y \frac{d u}{d y} + u + \frac{u}{\sqrt{u^{2} + 1}} = y

y \frac{d u}{d y} = y - u - \frac{u}{\sqrt{u^{2} + 1}}

(y - u - \frac{u}{\sqrt{u^{2} + 1}}) \frac{d y}{d u} = y

This belongs to an Abel equation of the second kind.
Let

v = y - u - \frac{u}{\sqrt{u^{2} + 1}}

Then

y = v + u + \frac{u}{\sqrt{u^{2} + 1}}

\frac{d y}{d u} = \frac{d v}{d u} + 1 + \frac{1}{{(u^{2} + 1)}^{\frac{3}{2}}}

∴ v (\frac{d v}{d u} + 1 + \frac{1}{{(u^{2} + 1)}^{\frac{3}{2}}}) = v + u + \frac{u}{\sqrt{u^{2} + 1}}

v \frac{d v}{d u} + (1 + \frac{1}{{(u^{2} + 1)}^{\frac{3}{2}}}) v = v + u + \frac{u}{\sqrt{u^{2} + 1}}

v \frac{d v}{d u} = - \frac{v}{{(u^{2} + 1)}^{\frac{3}{2}}} + u + \frac{u}{\sqrt{u^{2} + 1}}

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