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  • Solve the differential equation <img alt="x\cos \left ( \frac{y}{x} \right )\frac{dy}{dx}= y\cos \left ( \frac{y}{x} \right )+x\cdot" src="https://entrancecorner.codecogs.com/gif.latex?x%5Ccos%20%5Cleft%20%28%20%5Cfrac%7By%7D%7Bx%7D%20%5Cright%20%29%5C

Solve the differential equation x\cos \left ( \frac{y}{x} \right )\frac{dy}{dx}= y\cos \left ( \frac{y}{x} \right )+x\cdot

 

 

 

 
 
 
 
 

Answers (1)

(given) x\cos \left ( \frac{y}{x} \right )\frac{dy}{dx}= y\cos \left ( \frac{y}{x} \right )+x---(i)
which is a Homogeneous D.E On putting y = vx
\frac{dy}{dx}= v+x\frac{dv}{dx}  in equation (1) we get
x\cos v\left [ v+x\frac{dv}{dx} \right ]= vx\cos v+x
\Rightarrow v+\frac{xdv}{dx}= \frac{\left ( v\cos v+1 \right )}{\cos v}
\Rightarrow x\frac{dv}{dx}= \frac{v\cos v+1}{\cos v}-v
\Rightarrow x\frac{dv}{dx}= \frac{v\cos v+1-v\cos v}{\cos v}
\Rightarrow x\frac{dv}{dx}= \frac{1}{\cos v}\Rightarrow \cos vdv= \frac{dx}{x}
On Integrating both sides,  we get
\int \cos vdv= \int \frac{dx}{x}
\Rightarrow \sin v= \log x+c
\Rightarrow \sin \left ( \frac{y}{x} \right )= \log x+c      \left [ \because y= vx\Rightarrow v= \frac{y}{x} \right ]
which is the required solution of the Differential Equation
 

Posted by

Ravindra Pindel

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