(xdy)/(dx)=y+sqrt(x^2-y^2) — step-by-step solution

⚠ (xdy)/(dx)=y+sqrt(x^2-y^2) is read as (x)*y'=y+sqrt(x^2-y^2): use dif(expression, variable) for a derivative with respect to that variable, use (numerator)/(denominator) for a quotient

1) Let's solve the differential equation

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

, where

y

is a function of

x

; on an interval where

x > 0

, isolate

y'

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

On this interval, the right side only depends on

\frac{y}{x}

: the equation is homogeneous. With

v = \frac{y}{x}

, it reads

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

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' = v + \sqrt{1 - v^{2}}

, that is

x \cdot v' = \sqrt{1 - v^{2}}

The right side is defined for

v \in [-1, 1]

; this condition remains necessary for each branch

Before dividing,

\sqrt{1 - v^{2}} = 0

gives the equilibrium solutions

v = {-1, 1}

, that is

(y = - x \lor y = x)

On intervals where

\sqrt{1 - v^{2}} \neq 0

, separate the variables:

\frac{dv}{\sqrt{1 - v^{2}}} = \frac{dx}{x}

⇥1.1) Let's integrate:

\int \frac{1}{\sqrt{1 - v^{2}}} d v

⇥ⓘ The derivative of

\arcsin (v)

\frac{1}{\sqrt{1 - v^{2}}}

, so

\arcsin (v)

is an antiderivative of

\frac{1}{\sqrt{1 - v^{2}}}

\int \frac{1}{\sqrt{1 - v^{2}}} d v

\arcsin (v)

⇥

\arcsin (v)

Integrating both sides, with

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

since

x \neq 0

\arcsin (v) = \ln | x | + C

, where

C

is an arbitrary real constant, chosen independently on each interval

Isolate

v

v = \sin (\ln | x | + C)

. Since

y = v x

, the solution branches are:

These branches hold on intervals where

x > 0

. At points where

\sqrt{1 - v^{2}} = 0

, a join with an equilibrium must preserve the value and derivative and satisfy the original equation

On an interval where

x < 0

, isolate

y'

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

On this interval, the right side only depends on

\frac{y}{x}

: the equation is homogeneous. With

v = \frac{y}{x}

, it reads

y' = v - \sqrt{1 - v^{2}}

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' = v - \sqrt{1 - v^{2}}

, that is

x \cdot v' = - \sqrt{1 - v^{2}}

The right side is defined for

v \in [-1, 1]

; this condition remains necessary for each branch

Before dividing,

- \sqrt{1 - v^{2}} = 0

gives the equilibrium solutions

v = {-1, 1}

, that is

(y = - x \lor y = x)

On intervals where

- \sqrt{1 - v^{2}} \neq 0

, separate the variables:

- \frac{dv}{\sqrt{1 - v^{2}}} = \frac{dx}{x}

⇥1.2) Let's compute the integral:

\int (- \frac{1}{\sqrt{1 - v^{2}}}) d v

⇥ⓘ The derivative of

\arccos (v)

- \frac{1}{\sqrt{1 - v^{2}}}

, so

\arccos (v)

is an antiderivative of

- \frac{1}{\sqrt{1 - v^{2}}}

\int (- \frac{1}{\sqrt{1 - v^{2}}}) d v

\arccos (v)

⇥

\arccos (v)

Integrating both sides, with

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

since

x \neq 0

\arccos (v) = \ln | x | + C

, where

C

is an arbitrary real constant, chosen independently on each interval

Isolate

v

v = \cos (\ln | x | + C)

. Since

y = v x

, the solution branches are:

These branches hold on intervals where

x < 0

. At points where

- \sqrt{1 - v^{2}} = 0

, a join with an equilibrium must preserve the value and derivative and satisfy the original equation

▶Solution:

((x > 0 \land (y = - x \lor y = x \lor y = x \cdot \sin (\ln | x | + C))) \lor (x < 0 \land (y = - x \lor y = x \lor y = x \cdot \cos (\ln | x | + C))))