Question

(x + y) dx + (y - x) dy = 0​

Answer 1

y=vx then differentiate it wrt x put both the values of dy/dx and y/x in the given eqn nd then solve it

Answer 2

Answer:

$\large\bold\red{2 { \tan }^{ - 1} x - ln( {x}^{2} + {y}^{2} ) = 0}$

Step-by-step explanation:

Given,

A differential equation

• $(x + y)dx + (y - x)dy = 0$

Further solving it,

We get,

$= > (x + y)dx = (x - y)dy \\ \\ = > \frac{dy}{dx} = \frac{x + y}{x - y}$

Now,

Dividing both Numerator and Denominator by x on RHS,

We get,

$= > \frac{dy}{dx} = \frac{1 + \frac{y}{x} }{1 - \frac{y}{x} } \: \: \: \: \: ...........(i)$

Now,

Let's assume that,

$y = vx$

Differentiating both sides wrt x,

We get,

$= > \frac{dy}{dx} = v + x \frac{dv}{dx}$

Therefore,

Substituting these values in eqn (i),

We get,

$= > v + x \frac{dv}{dx} = \frac{1 + v}{1 - v} \\ \\ = > x \frac{dv}{dx} = \frac{1 + v}{1 - v} - v \\ \\ = > x \frac{dv}{dx} = \frac{1 + v - v + {v}^{2} }{1 - v} \\ \\ = > x \frac{dv}{dx} = \frac{1 + {v}^{2} }{1 - v} \\ \\ = > \frac{1 - v}{1 + {v}^{2} } dv = \frac{1}{x} dx \\ \\ = > \frac{1}{1 + {v}^{2} } dv - \frac{v}{1 + {v}^{2} } dv = \frac{1}{x} dx \\ \\ = > \frac{1}{1 + {v}^{2} } dv - \frac{1}{2} \times \frac{2v}{1 + {v}^{2} } dv = \frac{1}{x} dx$

Integrating both sides,

We get,

$\int \frac{1}{1 + {v}^{2} } dv - \frac{1}{2} \int \frac{2v}{1 + {v}^{2} } dv = \int \frac{1}{x} dx$

But,

We know that,

• $\int \frac{1}{1 + {x}^{2} } dx = { \tan }^{ - 1} x$

• $\int \frac{2x}{1 + {x}^{2} }dx = ln(1 + {x}^{2} )$

• $\int \frac{1}{x} dx = ln(x)$

Therefore,

Substituting the values,

We get,

$= > { \tan}^{ - 1} v - \frac{ ln( 1 + {v}^{2} ) }{2} = ln(x)$

Further,

Substituting the values of v,

We get,

$= > { \tan }^{ - 1} \frac{y}{x} - \frac{ ln(1 + {( \frac{y}{x}) }^{2} ) }{2} = ln(x) \\ \\ = > 2{ \tan }^{ - 1 } \frac{y}{x} - ln( \frac{ {x}^{2} + {y}^{2} }{ {x}^{2} } ) = 2 ln(x) \\ \\ = > 2 { \tan }^{ - 1} \frac{y}{x} = 2 ln(x) + ln( \frac{ {x}^{2} + {y}^{2} }{ {x}^{2} } )$

But,

We know that,

• $b \: ln(a) = ln( {a}^{b} )$

• $ln(a) + ln(b) = ln(ab)$

Therefore,

We get,

$= > 2 { \tan}^{ - 1} x = ln( {x}^{2} ) + ln( \frac{ {x}^{2} + {y}^{2} }{ {x}^{2} } ) \\ \\ = > 2 { \tan }^{ - 1} x = ln( {x}^{2} \times \frac{ {x}^{2} + {y}^{2} }{ {x}^{2} } ) \\ \\ = > 2 { \tan }^{ - 1} x = ln( {x}^{2} + {y}^{2} ) \\ \\ = > 2 { \tan }^{ - 1} x - ln( {x}^{2} + {y}^{2} ) = 0$

Hence,

Required Solution of this DE is,

• $\large\bold{2 { \tan }^{ - 1} x - ln( {x}^{2} + {y}^{2} ) = 0}$