Question

Find a particular solution of the differential equation,
$\dfrac{dy}{dx}+y\cot x=4x\operatorname{cosec}x\,,\left( x\ne 0 \right)$, it is given that $y=0$ when $x=\dfrac{\pi }{2}$.

Answer 1

To solve this differential equation we need to know how to solve the linear differential equation of the form $\dfrac{dy}{dx}+Py=Q$. Its solution is given by $y\left( IF \right)=\int{\left( Q\times IF \right)dx+c}$ where IF is the integration factor and is given by the $IF={{e}^{\int{\left( Pdx \right)}}}$.

Complete step by step answer:
We are given the differential equation,
$\dfrac{dy}{dx}+y\cot x=4x\operatorname{cosec}x$
And we know that whenever differential equation is in the linear form i.e. $\dfrac{dy}{dx}+Py=Q$ then it’s direct solution is given by $y\left( IF \right)=\int{\left( Q\times IF \right)dx+c}$ where IF is the integration factor and is given by the $IF={{e}^{\int{\left( Pdx \right)}}}$.
So in the above question we have,
P = $\cot x$ and Q = $4x\operatorname{cosec}x$
Hence integration factor we will get as,

& IF={{e}^{\int{\left( Pdx \right)}}} \\\ & IF={{e}^{\int{\cot xdx}}} \\\ & IF={{e}^{\log \left( \sin x \right)}} \\\ \end{aligned}$$ And we know that $${{e}^{\log x}}=x$$, hence $IF=\sin x$ Now, for solution we will solve as, $y\left( IF \right)=\int{\left( Q\times IF \right)dx+c}$ $$y\sin x=\int{4x\operatorname{cosec}x\sin xdx}+c$$ We know that $\operatorname{cosec}x=\dfrac{1}{\sin x}\,so,\,\operatorname{cosec}x.\sin x=1$, hence $\begin{aligned} & y\sin x=\int{4xdx}+c \\\ & y\sin x=\dfrac{4{{x}^{2}}}{2}+c \\\ & y\sin x=2{{x}^{2}}+c\,\,\,\,\,....\left( 1 \right) \\\ \end{aligned}$ It is given that $y=0$ when $x=\dfrac{\pi }{2}$. So putting $$y=0\,and\,x=\dfrac{\pi }{2}$$ in the equation 1 we get, $\begin{aligned} & 0\times \sin \dfrac{\pi }{2}=2{{\left( \dfrac{\pi }{2} \right)}^{2}}+c \\\ & c=-2{{\left( \dfrac{\pi }{2} \right)}^{2}} \\\ & c=-\dfrac{{{\pi }^{2}}}{2} \\\ \end{aligned}$ Now putting back value of c in equation 1 we will get the particular solution of the differential equation, $\dfrac{dy}{dx}+y\cot x=4x\operatorname{cosec}x\,,\left( x\ne 0 \right)$ as, **$\begin{aligned} & y\sin x=2{{x}^{2}}+c\, \\\ & y\sin x=2{{x}^{2}}-\dfrac{{{\pi }^{2}}}{2} \\\ \end{aligned}$** **Note:** While solving the problem many students may make mistake while assuming the values of P and Q from the given expression and may interchange both of their values so first you should write down the equation $\dfrac{dy}{dx}+Py=Q$ and then compare the given differential equation that if it is of this type then proceed further and solve otherwise try something else and also while calculating the integration factor, you need to be cautious because many students interchange P and Q there also hence will get wrong integration factor so wrong answer will be obtained.