Question

If $\sin x$ is the integrating factor of the linear differential equation $\dfrac{{dy}}{{dx}} + Py = Q$, then $P$ is ?

Answer 1

The integration of a differential equation can be made easier by multiplying it with the integrating factor. We will use the integrating factor formula i.e., $I.F = {e^{\int {Pdx} }}$, where $P$ is the function of $x$ and I.F is the integrating factor. Then we will find the value of $P$ by equating the I.F to $\sin x$, i.e., ${e^{\int {Pdx} }} = \sin x$.

Complete step by step answer:
For the differential equation given above i.e., $\dfrac{{dy}}{{dx}} + Py = Q$, we know that the integrating factor is of the form $I.F = {e^{\int {Pdx} }}$. Here $P$ and $Q$ are functions of $x$. But the given I.F is $\sin x$. So, we will equate ${e^{\int {Pdx} }}$ to $\sin x$. Hence, we get: ${e^{\int {Pdx} }} = \sin x$

Since $P$ is in the exponential of $e$ and we have to find the value of $P$, so in order to solve the equation we will take the logarithms of both sides L.H.S and R.H.S. So, the equation now becomes
$\Rightarrow \log {e^{\int {Pdx} }} = \log \sin x$........(The base of the $\log$ is $e$)
On further simplification using the property of log we will get:
$\Rightarrow \int {Pdx = \log \sin x}$........(Since ${\log _e}e = 1$)

Now we know that the derivative of an integral is the function itself. So, for further simplification we will differentiate both sides of the above equation.
So, the equation now becomes:
$\Rightarrow \dfrac{{d\int {Pdx} }}{{dx}} = \dfrac{{d(\log \sin x)}}{{dx}}$
$\Rightarrow P = \dfrac{{d(\log \sin x)}}{{dx}}$

Now we know that $\dfrac{{d(\log x)}}{{dx}} = \dfrac{1}{x}$, so using this property and the chain rule of differentiation we will solve the above equation. So, the equation now becomes:
$\Rightarrow P = \dfrac{1}{{\sin x}} \times \dfrac{{d(\sin x)}}{{dx}}$
$\Rightarrow P = \dfrac{1}{{\sin x}} \times \cos x$
$\therefore P = \cot x$

Therefore, the value of $P$ is $\cot x$.

Note: Here one important point that we need to keep in mind in order to solve the question is the formula for integrating factors. The integrating factor method can be used to solve differential equations of first order in a very simple way. The given differential equation becomes easily solvable when we multiply it with the integrating factor.