{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.


\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)


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 

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