xyy^'=y^2+xsqrt(4x^2+y^2) — step-by-step solution

⚠ xyy^'=y^2+xsqrt(4x^2+y^2) is read as x*y*y'=y^2+x*sqrt(4x^2+y^2): write y' or y'' for a derivative, use * for multiplication

Let's solve the differential equation

x y \cdot y' = y^{2} + x \cdot \sqrt{4 x^{2} + y^{2}}

, where

y

is a function of

x

; on an interval where

x > 0

and

y \neq 0

, isolate

y'

y' = \frac{x \cdot \sqrt{4 x^{2} + y^{2}} + y^{2}}{x y}

On this interval, the right side only depends on

\frac{y}{x}

: the equation is homogeneous. With

v = \frac{y}{x}

, it reads

y' = \frac{v^{2} + \sqrt{v^{2} + 4}}{v}

Let

y = v x

, where

v

is a function of

x

: then

y' = v + x \cdot v'

, and the equation becomes

v + x \cdot v' = \frac{v^{2} + \sqrt{v^{2} + 4}}{v}

, that is

x \cdot v' = \frac{\sqrt{v^{2} + 4}}{v}

\frac{\sqrt{v^{2} + 4}}{v}

is never zero: separate the variables,

\frac{v}{\sqrt{v^{2} + 4}} \cdot dv = \frac{dx}{x}

Let

u = v^{2} + 4

, so

u' = 2 v

. Since

u > 0

, using

\frac{d}{d v} (\sqrt{u}) = \frac{u'}{2 \sqrt{u}}

gives

\frac{d}{d v} (\sqrt{v^{2} + 4}) = \frac{2 v}{2 \cdot \sqrt{v^{2} + 4}}

\frac{v}{\sqrt{v^{2} + 4}}

, hence

\int \frac{v}{\sqrt{v^{2} + 4}} d v = \sqrt{v^{2} + 4}

Integrating both sides, with

\int \frac{1}{x} d x = \ln | x |

since

x \neq 0

\sqrt{v^{2} + 4} = \ln | x | + C

, where

C

is an arbitrary real constant, chosen independently on each interval

Replace

v

\frac{y}{x}

\sqrt{\frac{y^{2}}{x^{2}} + 4} = \ln | x | + C

On an interval where

x < 0

and

y \neq 0

, isolate

y'

y' = \frac{x \cdot \sqrt{4 x^{2} + y^{2}} + y^{2}}{x y}

On this interval, the right side only depends on

\frac{y}{x}

: the equation is homogeneous. With

v = \frac{y}{x}

, it reads

y' = \frac{v^{2} - \sqrt{v^{2} + 4}}{v}

Let

y = v x

, where

v

is a function of

x

: then

y' = v + x \cdot v'

, and the equation becomes

v + x \cdot v' = \frac{v^{2} - \sqrt{v^{2} + 4}}{v}

, that is

x \cdot v' = - \frac{\sqrt{v^{2} + 4}}{v}

- \frac{\sqrt{v^{2} + 4}}{v}

is never zero: separate the variables,

- \frac{v}{\sqrt{v^{2} + 4}} \cdot dv = \frac{dx}{x}

First take the factor

-1

outside the integral:

\int (- \frac{v}{\sqrt{v^{2} + 4}}) d v

- \int \frac{v}{\sqrt{v^{2} + 4}} d v

Let

u = v^{2} + 4

, so

u' = 2 v

. Since

u > 0

, using

\frac{d}{d v} (\sqrt{u}) = \frac{u'}{2 \sqrt{u}}

gives

\frac{d}{d v} (\sqrt{v^{2} + 4}) = \frac{2 v}{2 \cdot \sqrt{v^{2} + 4}}

\frac{v}{\sqrt{v^{2} + 4}}

, hence

\int \frac{v}{\sqrt{v^{2} + 4}} d v = \sqrt{v^{2} + 4}

; therefore

\int (- \frac{v}{\sqrt{v^{2} + 4}}) d v = - \sqrt{v^{2} + 4}

Integrating both sides, with

\int \frac{1}{x} d x = \ln | x |

since

x \neq 0

- \sqrt{v^{2} + 4} = \ln | x | + C

, where

C

is an arbitrary real constant, chosen independently on each interval

Replace

v

\frac{y}{x}

- \sqrt{\frac{y^{2}}{x^{2}} + 4} = \ln | x | + C

▶Solution:

((x > 0 \land \sqrt{\frac{y^{2}}{x^{2}} + 4} = \ln | x | + C \land y \neq 0) \lor (x < 0 \land - \sqrt{\frac{y^{2}}{x^{2}} + 4} = \ln | x | + C \land y \neq 0))