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प्रश्न
Solve the following differential equation:- \[x \cos\left( \frac{y}{x} \right)\frac{dy}{dx} = y \cos\left( \frac{y}{x} \right) + x\]
उत्तर
We have,
\[x \cos \left( \frac{y}{x} \right)\frac{dy}{dx} = y \cos\left( \frac{y}{x} \right) + x\]
\[ \Rightarrow \frac{dy}{dx} = \frac{y \cos\left( \frac{y}{x} \right) + x}{x \cos \left( \frac{y}{x} \right)} . . . . . \left( 1 \right)\]
Clearly this is a homogeneous equation,
Putting y = vx
\[ \Rightarrow \frac{dy}{dx} = v + x\frac{dv}{dx}\]
\[\text{Substituting }y = vx\text{ and }\frac{dy}{dx} = v + x\frac{dv}{dx}\text{ in (1) we get}\]
\[\frac{dy}{dx} = \frac{y \cos\left( \frac{y}{x} \right) + x}{x \cos \left( \frac{y}{x} \right)}\]
\[ \Rightarrow v + x\frac{dv}{dx} = \frac{vx \cos \left( v \right) + x}{x \cos \left( v \right)}\]
\[ \Rightarrow v + x\frac{dv}{dx} = \frac{v \cos \left( v \right) + 1}{\cos \left( v \right)}\]
\[ \Rightarrow x\frac{dv}{dx} = \frac{v \cos \left( v \right) + 1}{\cos \left( v \right)} - v\]
\[ \Rightarrow x\frac{dv}{dx} = \frac{v \cos \left( v \right) + 1 - v \cos \left( v \right)}{\cos \left( v \right)}\]
\[ \Rightarrow x\frac{dv}{dx} = \frac{1}{\cos \left( v \right)}\]
\[ \Rightarrow \cos \left( v \right) dv = \frac{1}{x}dx\]
Integrating both sides, we get
\[\int\cos \left( v \right) dv = \int\frac{1}{x}dx\]
\[ \Rightarrow \sin \left( v \right) = \log \left| x \right| + \log \left| C \right|\]
\[ \Rightarrow \sin \left( \frac{y}{x} \right) = \log \left| Cx \right|\]
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