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.