Question

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

Answer 1

First of all write $y=vx$ where v is also a variable then differentiate on both the sides with respect to x and substitute this value of $\dfrac{dy}{dx}$ and $\dfrac{y}{x}=v$ in the above differential equation. Then simplify this differential equation and during simplification you have to integrate the expression to get the solution.

Complete step by step answer:
The differential equation given in the above problem is as follows:
$x\dfrac{dy}{dx}-y+x\sin \left( \dfrac{y}{x} \right)=0$…………. Eq. (1)
Now, we have to find the particular solution of this equation.
Let us assume that:
$y=vx$……… Eq. (2)
In the above equation, v is also a variable so while differentiation we have to differentiate v also. Now, differentiating both the sides with respect to x we get,
$\begin{aligned} & \dfrac{dy}{dx}=v\left( 1 \right)+x\dfrac{dv}{dx} \\\ & \Rightarrow \dfrac{dy}{dx}=v+x\dfrac{dv}{dx} \\\ \end{aligned}$
We have differentiated $y=vx$ by using the product rule.
Product rule is applied on the two terms in which we write the first term as it is and differentiate the second term then add this to the differentiation of first term followed by multiplication of the second term which is written as it is.

Rewriting the equation (2) we get,
$\dfrac{y}{x}=v$
Now, substituting $\dfrac{y}{x}=v$ and $\dfrac{dy}{dx}$ that we have got above in eq. (1) we get,
$\begin{aligned} & x\left( v+x\dfrac{dv}{dx} \right)-vx+x\sin \left( v \right)=0 \\\ & \Rightarrow vx+{{x}^{2}}\dfrac{dv}{dx}-vx+x\sin \left( v \right)=0 \\\ \end{aligned}$
In the above equation, vx will be cancelled out and we get,
${{x}^{2}}\dfrac{dv}{dx}+x\sin \left( v \right)=0$
Taking x as common from the above equation we get,
$x\left( x\dfrac{dv}{dx}+\sin v \right)=0$ ………. Eq. (3)
Equating the expression written in the bracket and x to 0 we get,

& \left( x\dfrac{dv}{dx}+\sin v \right)=0; \\\ & x=0 \\\ \end{aligned}$$ In the above equations, we are rejecting $x=0$ because we have to find a solution for $x=2$. Now, solving the differential equation we get, $$\begin{aligned} & \left( x\dfrac{dv}{dx}+\sin v \right)=0 \\\ & \Rightarrow x\dfrac{dv}{dx}=-\sin v \\\ \end{aligned}$$ Writing v term with dv and x term with dx we get, $-\dfrac{dv}{\sin v}=\dfrac{dx}{x}$ Integrating on both the sides we get, $\begin{aligned} & \int{-\dfrac{dv}{\sin v}}=\int{\dfrac{dx}{x}} \\\ & \Rightarrow -\int{\cos ecvdv=\int{\dfrac{dx}{x}}} \\\ \end{aligned}$ $\begin{aligned} & \Rightarrow -\left( -\ln \left( \cos ecv+\cot v \right) \right)=\ln x+c \\\ & \Rightarrow \ln \left( \cos ecv+\cot v \right)=\ln x+c \\\ \end{aligned}$ Taking exponential on both the sides we get, $\begin{aligned} & \cos ecv+\cot v={{e}^{\ln x+c}} \\\ & \Rightarrow \cos ecv+\cot v=x.{{e}^{c}} \\\ \end{aligned}$ Substituting $v=\dfrac{y}{x}$ in the above equation we get, $\cos ec\left( \dfrac{y}{x} \right)+\cot \left( \dfrac{y}{x} \right)=x.{{e}^{c}}$………… Eq. (4) Now, we have to find the solution at $x=2,y=\pi $ by substituting these values in the above equation we get, $\begin{aligned} & \cos ec\left( \dfrac{\pi }{2} \right)+\cot \left( \dfrac{\pi }{2} \right)=2.{{e}^{c}} \\\ & \Rightarrow 1+0=2.{{e}^{c}} \\\ & \Rightarrow \dfrac{1}{2}={{e}^{c}} \\\ \end{aligned}$ Substituting the value of ${{e}^{c}}=\dfrac{1}{2}$ in eq. (4) we get, $\cos ec\left( \dfrac{y}{x} \right)+\cot \left( \dfrac{y}{x} \right)=x.\left( \dfrac{1}{2} \right)$ Rewriting the above equation we get, $\begin{aligned} & \dfrac{1}{\sin \left( \dfrac{y}{x} \right)}+\dfrac{\cos \left( \dfrac{y}{x} \right)}{\sin \left( \dfrac{y}{x} \right)}=x.\left( \dfrac{1}{2} \right) \\\ & \Rightarrow \dfrac{1+\cos \left( \dfrac{y}{x} \right)}{\sin \left( \dfrac{y}{x} \right)}=x\left( \dfrac{1}{2} \right) \\\ & \Rightarrow 2\left( 1+\cos \left( \dfrac{y}{x} \right) \right)=x\sin \left( \dfrac{y}{x} \right) \\\ \end{aligned}$ **Hence, we have got the particular solution of the differential equation as: $2\left( 1+\cos \left( \dfrac{y}{x} \right) \right)=x\sin \left( \dfrac{y}{x} \right)$** **Note:** Don’t accept all the solutions blindly such as in the above solution in eq. (3) two solutions are possible but we have rejected the solution $x=0$ because in the solution, we have to find the particular solution at $x=2$. The following is the eq. (3): $x\left( x\dfrac{dv}{dx}+\sin v \right)=0$ And by equating x and the expression written in brackets to 0 we get two solutions. $$\begin{aligned} & \left( x\dfrac{dv}{dx}+\sin v \right)=0; \\\ & x=0 \\\ \end{aligned}$$ And we have rejected $x=0$ solution because we have to find a solution at $x=2$. One more thing while integrating on both the sides of the equation, don’t forget to add some constant after integration. For instance, in the above solution we have added constant c as follows: $\begin{aligned} & \int{-\dfrac{dv}{\sin v}}=\int{\dfrac{dx}{x}} \\\ & \Rightarrow -\int{\cos ecvdv=\int{\dfrac{dx}{x}}} \\\ \end{aligned}$ $\begin{aligned} & \Rightarrow -\left( -\ln \left( \cos ecv+\cot v \right) \right)=\ln x+c \\\ & \Rightarrow \ln \left( \cos ecv+\cot v \right)=\ln x+c \\\ \end{aligned}$ If you forget to add this constant c then you cannot get the factor of $\dfrac{1}{2}$ which we found by substituting the point $x=2,y=\pi $ in the above equation.