Question

What is the particular solution of the differential equation $y' + y\tan x = \sin \left( {2x} \right)$ where $y\left( 0 \right) = 1$?

Answer 1

Hint: Here, in the given question, we are given differential equation $y' + y\tan x = \sin \left( {2x} \right)$ where $y\left( 0 \right) = 1$ and we need to find the particular solution of the differential equation. To find the answer to the given equation, first we have to observe the type of given differential equation, which is the linear differential equation $\dfrac{{dy}}{{dx}} + Py = Q$. Then we will find its integrating factor as ${e^{\int {Pdx} }}$ and find a particular solution of the equation by formula by substituting constant.

Complete step by step answer:
We have, $y' + y\tan x = \sin \left( {2x} \right)$
This can also be written as:
$\dfrac{{dy}}{{dx}} + \tan x.y = \sin \left( {2x} \right).........\left( i \right)$
This is a first order linear differential equation of the form:
$\dfrac{{dy}}{{dx}} + Py = Q$
where $P$ and $Q$ are functions of $x$.

This type of differential equations are solved when they are multiplied by a factor, which is called an integrating factor, because by multiplication of this factor the left hand side of the differential equation $\left( i \right)$ becomes the exact differential of some function.For the given equation,
$y' + \tan x.y = \sin \left( {2x} \right)$, $P = \tan x$ and $Q = \sin \left( {2x} \right)$
We will solve this using an integrating factor.
$ \Rightarrow I.F. = {e^{\int {Pdx} }}$

On substituting value of $P$, we get
$ \Rightarrow I.F. = {e^{\int {\tan xdx} }}$
As we know
$ \Rightarrow I.F. = {e^{\log \sec x}}$
$ \Rightarrow I.F. = \sec x$
Multiplying both sides of $\left( i \right)$ by integrating factor $\left( {\sec x} \right)$, we get
$ \Rightarrow \sec x\dfrac{{dy}}{{dx}} + \sec x\tan x.y = \sec x\sin \left( {2x} \right)$
Integrating both sides with respect to $x$, we get
$ \Rightarrow y\left( {\sec x} \right) = \int {\left( {\sin 2x \times \sec x} \right)} dx + C$ [Using: $y\left( {I.F.} \right) = \int {\left( {Q \times I.F.} \right)} dx + C$]
As we know $\sec x = \dfrac{1}{{\cos x}}$ and $\sin 2x = 2\sin x\cos x$. Therefore, we get
\[ \Rightarrow y\dfrac{1}{{\cos x}} = \int {\dfrac{1}{{\cos x}}\left( {2\sin x\cos x} \right)dx} + C\]

On cancelling out common terms on RHS, we get
\[ \Rightarrow \dfrac{y}{{\cos x}} = \int {2\sin xdx} \]
\[ \Rightarrow \dfrac{y}{{\cos x}} = 2\int {\sin xdx} \]
As we know $\int {\sin x = - \cos x + C} $. Therefore, we get
$ \Rightarrow \dfrac{y}{{\cos x}} = - 2\cos x + C$
On cross multiplication, we get
$ \Rightarrow y = - 2{\cos ^2}x + C\cos x$
Which is the general solution. Then using $y\left( 0 \right) = 1$ we can find $C$ as:
$ \Rightarrow 1 = - 2{\cos ^2}0 + C\cos 0$
As we know $\cos 0 = 1$. Therefore, we get
$ \Rightarrow 1 = - 2 + C$
$ \Rightarrow C = 3$
Hence, $y = - 2{\cos ^2}x + 3\cos x$.

Therefore, the particular solution of the differential equation $y' + y\tan x = \sin \left( {2x} \right)$ where $y\left( 0 \right) = 1$ is $y = - 2{\cos ^2}x + 3\cos x$.

Note: While solving differential equations, firstly rearrange the differential equation in such a way that you can get a differential equation in some known form such as variable separable, homogeneous or linear differential equation. While finding a particular solution, finding the value of constant very accurately as a particular solution is a unique solution.