Question

Find the particular solution of the differential equation $\dfrac{dx}{dy}+x\cot y=2y+{{y}^{2}}\cot y,(y\ne 0),$ given that x = 0 when y = $\dfrac{\pi }{\begin{align}
  & 2 \\
 & \\
\end{align}}$ .

Answer 1

Hint: To find the answer to the given equation, first we have to observe the type of given differential equation, which is linear differential equation $\dfrac{dx}{dy}+Rx=s$. Then we will find its integrating factor as ${{e}^{\int{Rdy}}}$ and find particular solution of the equation by formula $x.I.F=\int{(s.I.F)dy}+c$ by substituting given values of x and y such that x = 0 when y = $\dfrac{\pi }{\begin{align}
  & 2 \\
 & \\
\end{align}}$.

Complete step-by-step solution:
The given equation is $\dfrac{dx}{dy}+x\cot y=2y+{{y}^{2}}\cot y,(y\ne 0)$ .
We have to find a particular solution of the given equation.
The given equation it is in the form of linear differential equation $\dfrac{dx}{dy}+Rx=s$ .
By comparing with the given equation, we get R = cot y and s = ${{y}^{2}}\cot y$
Now we will go step by step to find our particular solution.
The formula for finding the particular solution of the differential equation is $x.I.F=\int{(s.I.F)dy}+c...(1)$
So, first we will find integrating factor,
The formula for the integrating factor is I.F. = ${{e}^{\int{Rdy}}}$
Substituting R = cot y, we have
$I.F.={{e}^{\int{(\cot y)dy}}}$
Now, we know that $\int{\cot y\text{ }dy}=\log \sin y+C$ , so we get
$\begin{align}
  & I.F.={{e}^{\log (\sin y)}} \\
 & \because {{e}^{\log x}}=x \\
 & \Rightarrow I.F.=\sin y \\
\end{align}$
Thus, our integrating factor is $\sin y$.
Now we will substitute the value of integrating factor and s in the formula of particular solution which we have mentioned above as (1), we will get
$\begin{align}
  & x.\sin y=\int{(2y+{{y}^{2}}\cot y).\sin ydy+c} \\
 & \therefore x.\sin y=\int{(2y\sin ydy)+{{y}^{2}}\cot y(\sin y)dy.+c} \\
\end{align}$
Here, we know $\dfrac{\cos y}{\sin y}=\cot y$, we will substitute cot y in above expression, we get
$\begin{align}
  & x.\sin y=\int{\left( (2y\sin y)+{{y}^{2}}.\left( \dfrac{\cos y}{\sin y} \right)(\sin y) \right)dy.+c} \\
 & \\
\end{align}$
$\begin{align}
  & \therefore x.\sin y=\int{\left( (\sin y.2y)+{{y}^{2}}\cos y \right)dy.+c} \\
 & \\
\end{align}$
Now, we can apply sum rule, stated as: \[\int{(f(x)+g(x))dx=\int{f(x)dx+\int{g(x)}}}dx\]
So, our previous expression becomes,
$\therefore x.\sin y=\int{\sin y.2ydy+\int{{{y}^{2}}\cos ydy+c}}...(2)$
Now, we will evaluate the integrals separately.
Taking constant term out from first integral, we get
$\int{(\sin y).2ydy}=2\int{\sin y.ydy}$
We will use integration by parts rule for function $y=u.v$ , stated as $\int{u.vdy=u\int{vdy-\int{u'\int{(vdy}}}})dy$
As we know the concept of ILATE formula which helps us decide the function u and v, we will use it and assume u and v. First function, i.e u must be taken in the order as Inverse, Logarithmic, Algebraic, Trigonometric and Exponential function.
Following this, here we will take u = sin y and v = y, so we will get
\[2\int{\sin y.ydy=2\left[ \sin y\int{ydy-\int{\dfrac{dy}{dx}\left( \sin y \right)\int{\left( ydy \right)dy}}} \right]}\]
We know that $\int{{{x}^{n}}dx=\dfrac{{{x}^{n+1}}}{n+1}+C}\text{ and }\int{\sin xdx=\cos x+C}$ , so we get
\[\begin{align}
  & \Rightarrow 2\int{\sin y.ydy}=2\left[ \sin y\left( \dfrac{{{y}^{2}}}{2} \right)-\int{\left( \cos y \right).\left( \dfrac{{{y}^{2}}}{2} \right)}dy \right] \\
 & \Rightarrow 2\int{\sin y.ydy}=\sin y\left( {{y}^{2}} \right)-\int{\cos y.\left( {{y}^{2}} \right)}dy \\
 & \Rightarrow 2\int{\sin y.ydy}={{y}^{2}}\sin y-\int{{{y}^{2}}\cos y}dy \\
\end{align}\]
Now we will substitute above expression in (2), we get
$\Rightarrow x.\sin y={{y}^{2}}\sin y-\int{{{y}^{2}}\cos ydy}+\int{{{y}^{2}}\cos y}dy+c$
Cancelling similar terms, we get,
$\Rightarrow x.\sin y={{y}^{2}}\sin y+c...(3)$
We have the given conditions as x = 0 when y = $\dfrac{\pi }{2}$. So, we will substitute value of x and y in (3), and we will get
$\Rightarrow 0.\sin \left( \dfrac{\pi }{2} \right)={{\left( \dfrac{\pi }{2} \right)}^{2}}\sin \left( \dfrac{\pi }{2} \right)+c$
We know that $\begin{align}
  & \sin \left( \dfrac{\pi }{2} \right)=1 \\
 & \\
\end{align}$ . So, we get
$\begin{align}
  & \Rightarrow 0=\dfrac{{{\pi }^{2}}}{4}+c \\
 & \Rightarrow c=-\dfrac{{{\pi }^{2}}}{4} \\
\end{align}$
Now we will substitute for c in (3), and we will get,
\[\begin{align}
  & \Rightarrow x.\sin y={{y}^{2}}\sin y-\dfrac{{{\pi }^{2}}}{4} \\
 & \Rightarrow x={{y}^{2}}-\dfrac{{{\pi }^{2}}}{4}\operatorname{cosec}y \\
\end{align}\]
This is the required particular solution of the given equation.

Note: We can find solutions by observing the type of given equation quickly. We have to avoid confusion in choosing functions by using the ILATE rule while finding integral using the integration by parts method. Make sure you are using the formula for the particular solution $x.I.F=\int{(s.I.F)dy}+c$ and not getting this wrong. Here, we had found out $\int{\sin y.2ydy}$ first and then simplified. But, we can also find out $\int{{{y}^{2}}\cos ydy+c}$ first and then simplify, in both cases the integral terms would get canceled off and give us $x.\sin y={{y}^{2}}\sin y+c$.