{x cos y/x + y sin y/x}y dx = {y siny/x – x cos y/x}x dy

Question 3:

show that the given differential equation is homogeneous and solve it.

$\displaystyle\{x\cos (\frac{y}{x})+y\sin (\frac{y}{x})\}y dx = \{y\sin (\frac{y}{x}) - x\cos (\frac{y}{x})\}x dy$ .

Solution:

$\displaystyle\{x\cos (\frac{y}{x})+y\sin (\frac{y}{x})\}y dx = \{y\sin (\frac{y}{x}) - x\cos (\frac{y}{x})\}x dy$ .

The given differential equation can be expressed as

$\displaystyle \dfrac{dy}{dx} = \dfrac{\{x\cos (\frac{y}{x})+y\sin (\frac{y}{x})\}y}{\{y\sin (\frac{y}{x}) - x\cos (\frac{y}{x})\}x}$

$\displaystyle \dfrac{dy}{dx} = \dfrac{\{\cos (\frac{y}{x})+(\frac{y}{x})\sin (\frac{y}{x})\}(\frac{y}{x})}{\{(\frac{y}{x}) \sin (\frac{y}{x}) - \cos (\frac{y}{x})\} } = F(x, y)$

Now,

$F(\lambda x, \lambda y) = \dfrac{\{\cos (\frac{\lambda y}{\lambda x})+(\frac{\lambda y}{\lambda x})\sin (\frac{\lambda y}{\lambda x})\}(\frac{\lambda y}{\lambda x})}{\{(\frac{\lambda y}{\lambda x}) \sin (\frac{\lambda y}{\lambda x}) - \cos (\frac{\lambda y}{\lambda x})\}} = \lambda^0 F(x, y)$

Therefore, F(x, y) is a homogeneous function of degree zero. So, the given differential equation is a homogeneous differential equation .

Put y = vx

$\Rightarrow \dfrac{dy}{dx} = v + \dfrac{dv}{dx}$

Putting the value of dy/dx and y in the given equation, we have

$\displaystyle v + \dfrac{dv}{dx} = \dfrac{\{\cos (\frac{vx}{x})+(\frac{vx}{x})\sin (\frac{vx}{x})\}(\frac{vx}{x})}{\{(\frac{vx}{x}) \sin (\frac{vx}{x}) - \cos (\frac{vx}{x})\} }$

$\displaystyle \Rightarrow v + \dfrac{dv}{dx} = \dfrac{\{x\cos v + vx\sin v\} v x}{\{v x\sin v - x\cos v\}x}$

$\displaystyle \Rightarrow v + \dfrac{dv}{dx} = \dfrac{\{\cos v + v\sin v\} v }{\{v \sin v - \cos v\}}$

$\displaystyle \Rightarrow \dfrac{dv}{dx} = \dfrac{\{\cos v + v\sin v\} v }{\{v \sin v - \cos v\}}- v$

$\displaystyle \Rightarrow \dfrac{dv}{dx} = \dfrac{\cos v + v^2\sin v - v^2\sin v + v \cos v}{\{v \sin v - \cos v\}}$

$\displaystyle \Rightarrow \dfrac{dv}{dx} = \dfrac{2v \cos v }{\{v \sin v - \cos v\}}$

$\displaystyle \Rightarrow \dfrac{v \sin v - \cos v}{v\cos v} dv = \dfrac{2}{x} dx$

$\displaystyle \Rightarrow (\tan v - \frac{1}{v})dv = \dfrac{2}{x} dx$

$\displaystyle \Rightarrow \int (\tan v - \frac{1}{v})dv =2\int \dfrac{1}{x} dx$

$\displaystyle \Rightarrow \log|\sec v| - \log v = 2\log x + \log C$

$\displaystyle \Rightarrow \log|\sec v| - \log v - \log x^2 = \log C$

$\displaystyle \Rightarrow \log\dfrac{\sec v}{v x^2} = \log C$

$\displaystyle \Rightarrow \dfrac{\sec v}{v x^2} = C$

$\displaystyle \Rightarrow \dfrac{\sec (\frac{y}{x})}{\frac{y}{x} x^2} = C$

$\displaystyle \Rightarrow \sec(\frac{y}{x}) = Cxy$

Which is the required solution.

Some other question

Question 1: Solve the following differential equation: $(1 + e^{y/x})dy + e^{y/x}(1 - \frac{y}{x}) = 0,$ x ≠ 0

Solution : For solution click here

Question 2: Show that differential equation $\displaystyle 2ye^{x/y} dx + (y - 2xe^{x/y}) dy = 0$ is homogeneous and find its particular solution, given that x = 0 when y = 1.

Solution : For solution click here

{x cos y/x + y sin y/x}y dx = {y siny/x – x cos y/x}x dy

Question 3:

Solution:

Some other question

Team Gmath

2yxe^{x/y} dx + (y-2xe^{x/y} dx = 0

(1 + e^{y/x})dy + e^{y/x}(1 – \frac{y}{x}) = 0

A sum of ₹ 4,250 is to be used to give 10 cash prizes

A child puts one five-rupees coin of her savings in the

Yasmeen saves ₹ 32 during the first month ₹ 36

Show that the sum of all terms of an A.P. whose first term is a

Leave a Reply Cancel reply