Question

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Solve the differential equation below:

$\sf \large (x {}^{2} + y {}^{2} )dx - xy \: dy = 0.$
​

Answer 1

Question:-

Solve the differential equation below:

(x² + y²)dx – xy dy = 0

Given:-

• (x² + y²)dx – xy dy = 0.

To Solve:-

• The differential equation given below: (x² + y²)dx – xy dy = 0.

Solution:-

(x² + y²)dx – xy dy = 0.

⇒ (x² + y²)dx = xy dy.

$\sf \large ⇒ \frac{dy}{dx} = \frac{x {}^{2} + y {}^{2} }{xy} \: \: \: ...(i)$

Let, y = vx

$\sf \large \therefore \frac{dy}{dx} = v + x \frac{dv}{dx} \\ \\ \sf \large \therefore v + x \frac{dv}{dx} = \frac{x {}^{2} + v {}^{2} x {}^{2} }{x {}^{2} v} = \frac{1 + v {}^{2} }{v} \\ \\ \sf \large \therefore x \frac{dv}{dx} = \frac{1{{ \cancel{ + v {}^{2}}}}{{ \cancel{ - v {}^{2} }}}}{v} = \frac{1}{v} \\ \\ \sf \large \therefore xdv = \frac{1}{v} \: dx$

$\sf \large⇒ \int \frac{dx}{x} = \int v \: \: dv \\ \\ \sf \large⇒ log \: x = \frac{v {}^{2} }{2} + c \\ \\ \sf \large ⇒ log \: x = \frac{y {}^{2} }{2x {}^{2} } + c \\ \\ \sf \large⇒y {}^{2} + c = 2x {}^{2} log \: x \\ \\ \sf \large ⇒y {}^{2} = x {}^{2} log \: x {}^{2} c$

Answer:-

$\sf \large \color{red}⇒y {}^{2} = x {}^{2} log \: x {}^{2} c.$

Hope you have satisfied. ⚘

Answer 2

Question:-

Solve the differential equation below:

(x² + y²)dx – xy dy = 0

Given:-

(x² + y²)dx – xy dy = 0.

To Solve:-

The differential equation given below: (x² + y²)dx – xy dy = 0.

Solution:-

(x² + y²)dx – xy dy = 0.

⇒ (x² + y²)dx = xy dy.

\sf \large ⇒ \frac{dy}{dx} = \frac{x {}^{2} + y {}^{2} }{xy} \: \: \: ...(i)⇒

dx

dy

=

xy

x

2

+y

2

...(i)

Let, y = vx

\begin{gathered} \sf \large \therefore \frac{dy}{dx} = v + x \frac{dv}{dx} \\ \\ \sf \large \therefore v + x \frac{dv}{dx} = \frac{x {}^{2} + v {}^{2} x {}^{2} }{x {}^{2} v} = \frac{1 + v {}^{2} }{v} \\ \\ \sf \large \therefore x \frac{dv}{dx} = \frac{1{{ \cancel{ + v {}^{2}}}}{{ \cancel{ - v {}^{2} }}}}{v} = \frac{1}{v} \\ \\ \sf \large \therefore xdv = \frac{1}{v} \: dx\end{gathered}

∴

dx

dy

=v+x

dx

dv

∴v+x

dx

dv

=

x

2

v

x

2

+v

2

x

2

=

v

1+v

2

∴x

dx

dv

=

v

1

+v

2

−v

2

=

v

1

∴xdv=

v

1

dx

\begin{gathered} \sf \large⇒ \int \frac{dx}{x} = \int v \: \: dv \\ \\ \sf \large⇒ log \: x = \frac{v {}^{2} }{2} + c \\ \\ \sf \large ⇒ log \: x = \frac{y {}^{2} }{2x {}^{2} } + c \\ \\ \sf \large⇒y {}^{2} + c = 2x {}^{2} log \: x \\ \\ \sf \large ⇒y {}^{2} = x {}^{2} log \: x {}^{2} c\end{gathered}

⇒∫

x

dx

=∫vdv

⇒logx=

2

v

2

+c

⇒logx=

2x

2

y

2

+c

⇒y

2

+c=2x

2

logx

⇒y

2

=x

2

logx

2

c

Answer:-

\sf \large \color{red}⇒y {}^{2} = x {}^{2} log \: x {}^{2} c.⇒y

2

=x

2

logx

2

c.