Question

Solve the differential equation $\dfrac{{dy}}{{dx}} + y\cot x = 2x + {x^2}\cot x$

Answer 1

Hint: The given differential equation is a Bernoulli`s Equation of the form $\dfrac{{dy}}{{dx}} + Py = Q$ , which can be solved by using its Integrating factor. So, use this concept to reach the solution of the problem. For finding the integrating factor we used integration.

Complete step-by-step solution -
Given $\dfrac{{dy}}{{dx}} + y\cot x = 2x + {x^2}\cot x$
This differential equation is of the form
$\dfrac{{dy}}{{dx}} + Py = Q$
where P = \cot x{\text{ & }}Q = 2x + {x^2}\cot x
we know that $IF = {e^{\int {Pdx} }}$ , so for the given equation

$$\Rightarrow {\text{IF}} = {e^{\int {\cot xdx} }} \\\ \Rightarrow {\text{IF}} = {e^{\log x}}{\text{ }}\left[ {\because \int {\cot xdx = \log \left( {\sin x} \right)} } \right] \\\ \Rightarrow {\text{IF}} = \sin x{\text{ }}\left[ {\because {e^{\log x}} = x} \right] \\\ \therefore {\text{ IF }} = \sin x \\\$$

The solution of the differential equation of the form $\dfrac{{dy}}{{dx}} + Py = Q$ is given by $y \times {\text{IF }} = \int {\left( {Q \times {\text{IF}}} \right)} {\text{ }}dx + c$
So, for the given differential equation, the solution is

$$\Rightarrow y \times \sin x{\text{ }} = \int {\left[ {\left( {2x + {x^2}\cot x} \right) \times \sin x} \right]} {\text{ }}dx + c \\\ \Rightarrow y \times \sin x{\text{ }} = \int {\left( {2x\sin x + {x^2}\cot x\sin x} \right)dx + c} \\\ \Rightarrow y\sin x{\text{ }} = \int {2x\sin xdx} + \int {\left( {{x^2}\sin x \times \dfrac{{\cos x}}{{\sin x}}} \right)dx} + c \\\ \Rightarrow y\sin x = 2\int {x\sin xdx} + \int {{x^2}\cos xdx} + c \\\$$

By using the formula of integrating by parts
$\int f(x)\cdot g(x) dx = f(x)\int g(x)dx - \int (f’(x) \cdot (\int g(x) dx)) dx$
We have

\Rightarrow y\sin x = 2\left[ {\sin x\int {xdx - \int {\left\\{ {\dfrac{{d\left( {\sin x} \right)}}{{dx}}\int {xdx} } \right\\}dx} } } \right] + \int {{x^2}\cos xdx} + c \\\ \Rightarrow y\sin x = 2\left[ {\sin x \times \left( {\dfrac{{{x^2}}}{2}} \right) - \int {\left\\{ {\cos x \times \left( {\dfrac{{{x^2}}}{2}} \right)} \right\\}dx} } \right] + \int {{x^2}\cos xdx} + c \\\ \Rightarrow y\sin x = {x^2}\sin x - \int {{x^2}\cos xdx + \int {{x^2}\cos xdx} + c} \\\

Cancelling the common terms, we get
$\Rightarrow y\sin x = {x^2}\sin x + c$
Therefore, the solution of the differential equation $\dfrac{{dy}}{{dx}} + y\cot x = 2x + {x^2}\cot x$ is $y\sin x = {x^2}\sin x + c$.
Note: The integrating factor of the Bernoulli`s Equation of the form $\dfrac{{dy}}{{dx}} + Py = Q$ is given by $IF = {e^{\int {Pdx} }}$. We have used Integrating by parts method to solve the differential equation.