Question

(xtan y/x-ysec²y/x)dx+xsec²y/x dy=0​

Answer 1

Answer:

$\large\bold\red{c = x( \csc \frac{2y}{x} - \cot \frac{2y}{x} )}$

Step-by-step explanation:

Given,

A differential equation,

• $(x \tan \frac{y}{x} - y { \sec }^{2} \frac{y}{x} )dx + x { \sec }^{2} \frac{y}{x} dy = 0$

Further Solving,

We get,

$= > x { \sec }^{2} \frac{y}{x} dy =( y { \sec }^{2} \frac{y}{x} - x \tan \frac{y}{x} ) dx \\ \\ = > \frac{dy}{dx} = \frac{(y { \sec }^{2} \frac{y}{x} - x \tan \frac{y}{x} ) }{x { \sec }^{2} \frac{y}{x} } \\ \\ = > \frac{dy}{dx} = \frac{y}{x} - \sin \frac{y}{x} \cos \frac{y}{x} \\ \\ = > \frac{dy}{dx} = \frac{y}{x} - \frac{1}{2} \times 2 \sin( \frac{y}{ x }) \cos( \frac{y}{x} )$

But,

We know that,

• $2 \sin( \alpha ) \cos( \alpha ) = \sin(2 \alpha )$

Therefore,

We get,

$= > \frac{dy}{dx} = \frac{y}{x} - \frac{ \sin( \frac{2y}{x} ) }{2} \: \: \: \: \: .........(i)$

Now,

Let's assume that,

$y = vx$

Differentiating both sides wrt x,

We get,

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

Substituting these values in (i),

We get,

$= > v + x \frac{dv}{dx} = v - \frac{ \sin(2v) }{2} \\ \\ = > x \frac {dv}{dx} = - \frac{ \sin(2v) }{2} \\ \\ = > 2 \frac{dv}{ \sin(2v) } = - \frac{dx}{x} \\ \\ = > 2 \csc(2v) dv = - \frac{dx}{x}$

Now,

Integrating both the sides,

We get,

$= >2\int \csc(2v) dv = - \int \frac{dx}{x} \\ \\ = > 2 \frac{ ln | \csc(2v) - \cot(2v) | }{2} = - ln(x) + ln(c) \\ \\ = > ln( \csc(2v) - \cot(2v) ) + ln(x) = ln(c) \\ \\ = > ln(x( \csc2v - \cot2v ) ) = ln(c) \\ \\ = > x( \csc2v - \cot2v) = c$

Substituting the value of v,

We get,

The Solution of DE is,

$= > \large \bold{c = x( \csc \frac{2y}{x} - \cot \frac{2y}{x} )}$