Question

Find the particular solution of the given differential equation: $x\dfrac{dy}{dx}-y+x\sin \left( \dfrac{y}{x} \right)=0$, given that when $x=2$, $y=\pi $.

Answer 1

Hint: We start solving the problem by dividing both sides with x. We then assume $y=vx$ and convert $\dfrac{dy}{dx}$ in terms of $\dfrac{dv}{dx}$. We then substitute these terms in the differential equation and make necessary calculations to bring terms with similar variables on the same side. We then integrate both sides to get the general solution of the given differential equation. We then substitute $x=2$, $y=\pi $, and made necessary calculations to get the desired particular solution.

Complete step by step answer:
According to the problem, we are given a differential equation $x\dfrac{dy}{dx}-y+x\sin \left( \dfrac{y}{x} \right)=0$. We need to find the particular solution of this differential equation if it is given $x=2$, $y=\pi $.
Let us first find the general solution of the given differential equation and then find the particular solution using this general solution.
So, we have $x\dfrac{dy}{dx}-y+x\sin \left( \dfrac{y}{x} \right)=0$.
$\Rightarrow \dfrac{dy}{dx}-\dfrac{y}{x}+\dfrac{x}{x}\sin \left( \dfrac{y}{x} \right)=0$.
$\Rightarrow \dfrac{dy}{dx}-\dfrac{y}{x}+\sin \left( \dfrac{y}{x} \right)=0$.
$\Rightarrow \dfrac{dy}{dx}=\dfrac{y}{x}-\sin \left( \dfrac{y}{x} \right)$ ---(1).
Let us assume $y=vx$ ---(2).
Let us differentiate equation (2) on both sides with respect to x.
$\Rightarrow \dfrac{dy}{dx}=\dfrac{d}{dx}\left( vx \right)$.
We know that $\dfrac{d}{dx}\left( uv \right)=u\dfrac{dv}{dx}+v\dfrac{du}{dx}$.
$\Rightarrow \dfrac{dy}{dx}=v\dfrac{dx}{dx}+x\dfrac{dv}{dx}$.
We know that $\dfrac{dx}{dx}=1$.
$\Rightarrow \dfrac{dy}{dx}=v\left( 1 \right)+x\dfrac{dv}{dx}$.
$\Rightarrow \dfrac{dy}{dx}=v+x\dfrac{dv}{dx}$ ---(3).
Let us substitute equations (2) and (3) in equation (1).
$\Rightarrow v+\dfrac{dv}{dx}=\dfrac{vx}{x}-\sin \left( \dfrac{vx}{x} \right)$.
$\Rightarrow v+\dfrac{dv}{dx}=v-\sin \left( v \right)$.
$\Rightarrow \dfrac{dv}{dx}=v-v-\sin \left( v \right)$.
$\Rightarrow \dfrac{dv}{dx}=-\sin \left( v \right)$.
Let us bring the functions with similar variables on same side.
$\Rightarrow \dfrac{dv}{\sin \left( v \right)}=-dx$.
We know that $\dfrac{1}{\sin x}=\operatorname{cosec}x$.
$\Rightarrow \operatorname{cosec}vdv=-dx$.
Let us integrate on both sides.
$\Rightarrow \int{\operatorname{cosec}vdv}=-\int{dx}$.
We know that $\int{dx}=x+C$ and $\int{\operatorname{cosec}xdx=\ln \left( \operatorname{cosec}x-\cot x \right)}$.
$\Rightarrow \operatorname{cosec}v=-x+C$---(4), where C is an arbitrary constant.
From equation (2), we have $y=vx$.
$\Rightarrow v=\dfrac{y}{x}$. We substitute in this in equation (4).
$\Rightarrow \operatorname{cosec}\left( \dfrac{y}{x} \right)=-x+C$.
So, we have found the general solution of the given differential equation $x\dfrac{dy}{dx}-y+x\sin \left( \dfrac{y}{x} \right)=0$ as $\operatorname{cosec}\left( \dfrac{y}{x} \right)=-x+C$.
Let us substitute $x=2$, $y=\pi $ in the general solution in order to get the particular solution.
$\Rightarrow \operatorname{cosec}\left( \dfrac{\pi }{2} \right)=-2+C$.
$\Rightarrow 1=-2+C$.
$\Rightarrow 1+2=C$.
$\Rightarrow C=3$. Let us substitute this value in the general equation.
So, we get the particular solution as $\operatorname{cosec}\left( \dfrac{y}{x} \right)=-x+3$.
∴ We have found the particular solution for the differential equation $x\dfrac{dy}{dx}-y+x\sin \left( \dfrac{y}{x} \right)=0$ as $\operatorname{cosec}\left( \dfrac{y}{x} \right)=-x+3$.

Note:
Here we used $y=vx$ as the given function is homogeneous. We know that if a function $f\left( x,y \right)$ is said to be homogeneous with degree n, then it satisfies the property $f\left( vx,xy \right)={{v}^{n}}f\left( x,y \right)$. We can see that the given differential equation satisfies this which is shown below:
Let us assume $f\left( x,y \right)=x\dfrac{dy}{dx}-y+x\sin \left( \dfrac{y}{x} \right)$.
$\Rightarrow f\left( vx,vy \right)=vx\dfrac{d\left( vy \right)}{d\left( vx \right)}-vy+vx\sin \left( \dfrac{vy}{vx} \right)$.
$\Rightarrow f\left( vx,vy \right)=vx\dfrac{vdy}{vdx}-vy+vx\sin \left( \dfrac{y}{x} \right)$.
$\Rightarrow f\left( vx,vy \right)=vx\dfrac{dy}{dx}-vy+vx\sin \left( \dfrac{y}{x} \right)$.
$\Rightarrow f\left( vx,vy \right)=v\left( x\dfrac{dy}{dx}-y+x\sin \left( \dfrac{y}{x} \right) \right)$.
$\Rightarrow f\left( vx,vy \right)=vf\left( x,y \right)$.
We should not make calculation mistakes and confuse with the formulas of integration while solving this problem.