Question

solve xdy-ydx=cos(1/x)dx​

Answer 1

Given,

xdy-ydx = cos(1/x)dx​

To FInd

The solution of the given equation =?

Solution,

$xdy - ydx - cos(1/x)dx =0$

Dividing both sides by x²:

$xdy - ydx/x^2 - dx *cos(1/x)dx/x^2 =0\\xdy - ydx/x^2 = dx *cos(1/x)dx/x^2$

Let u = y / x

By differentiating u, we get

$du = (xdy - ydx)/x^2$ ⇒ Equation 1

Also, let p = 1 /x

By differentiating p, we get

$dp = - 1 / x^2$  ⇒ Equation 2

Putting values from equation 1 and 2, we get

⇒ du + (cos p) * dp = 0

By integrating both sides, we get

⇒ u + sin p + c

Putting vaue of u = y / x and p = 1/x

⇒ (y / x) + sin(1/x) + c

Hence, the solution of xdy-ydx=cos(1/x)dx​ is (y / x) + sin(1/x) + c.

Answer 2

Concept:-

This is about how to solve a differential equation.

Given:-

$xdy-ydx=cos(1/x)dx$

Find:-

We have to find the solution to the given equation.

Solution:-

According to the problem,

$xdy-ydx=cos(1/x)dx$

we can write it,

$x\frac{dy}{dx}-y=cos \frac{1}{x}$

now multiplying both sides by $\frac{1}{x^{2} }$

$\frac{1}{x} \frac{dy}{dx}-\frac{y}{x^{2} } =\frac{1}{x^{2} } cos \frac{1}{x}$

or,$\frac{d}{dx}\frac{y}{x}=\frac{1}{x^{2} } cos \frac{1}{x}$

or,$d\frac{y}{x}=\frac{1}{x^{2} } cos \frac{1}{x}dx$

integrate both sides,

$\int\limi d\frac{y}{x}=\int\limits \frac{1}{x^{2} } cos \frac{1}{x}dx$

or, $\frac{y}{x}=-\int\limits cos \frac{1}{x}d(\frac{1}{x} )$

or, $\frac{y}{x}=-sin \frac{1}{x} + c$

or,$\frac{y}{x} +sin \frac{1}{x} = c$

or,$y +xsin \frac{1}{x} = cx$

where c is a constant.

Hence, the $y +xsin \frac{1}{x} = cx$ is the solution of the given equation.