Question

Solve the differential equation
$\dfrac{dy}{dx}+y\tan x=\sin x$

Answer 1

Hint: Observe that the given differential equation is a linear differential equation, i.e. a differential equation of the form $\dfrac{dy}{dx}+P\left( x \right)y=Q\left( x \right)$. The solution of a linear differential equation is given by
$yI.F=\int{Q\left( x \right)I.Fdx}$, where $IF={{e}^{\int{P\left( x \right)dx}}}$. Find the integrating factor of the given differential equation and hence find the solution of the given equation. Verify your answer.

Complete step-by-step answer:
We have $\dfrac{dy}{dx}+y\tan x=\sin x$, which is of the form of $\dfrac{dy}{dx}+P\left( x \right)y=Q\left( x \right)$.
Here $P\left( x \right)=\tan x$ and $Q\left( x \right)=\sin x$.
Now, we know that the integrating factor IF of a linear differential equation is given by
$IF={{e}^{\int{P\left( x \right)dx}}}$
Hence, the integrating factor of the given differential equation is given by
$IF={{e}^{\int{\tan xdx}}}$.
We know that $\int{\tan xdx}=\ln \left( \sec x \right)+C$
Hence, we have $IF={{e}^{\ln \left( \sec x \right)}}$
We know that ${{e}^{\ln x}}=x$
Hence, we have
$IF=\sec x$
Now, we know that the solution of a linear differential equation is given by
$yIF=\int{Q\left( x \right)IFdx}$
Hence, the solution of the given equation is
$y\sec x=\int{\sin x\sec xdx}$
Now, we have
$\sin x\sec x=\dfrac{\sin x}{\cos x}=\tan x$
Hence, we have
$y\sec x=\int{\tan xdx}$
We know that
\[\int{\tan xdx}=\ln \left( \sec x \right)\]
Hence, we have
$y\sec x=\ln \left( \sec x \right)+C$
Dividing both sides by $\sec x$, we get
$y=\cos x\ln \left( \sec x \right)+C\cos x$, which is the required solution of the given differential equation.

Note: Verification.
We have
$y=\cos x\ln \left( \sec x \right)+C\cos x$
Differentiating both sides, we get
$\dfrac{dy}{dx}=-\sin x\ln \left( \sec x \right)+\cos x\tan x-C\sin x$
Adding $y\tan x$ on both sides, we get
$\dfrac{dy}{dx}+y\tan x=\dfrac{dy}{dx}=-\sin x\ln \left( \sec x \right)+\cos x\tan x-C\sin x+\tan x\left( \cos x\ln \left( \sec x \right)+C\cos x \right)$
We have $\tan x\cos x=\dfrac{\sin x}{\cos x}\cos x=\sin x$
Hence, we have
$\dfrac{dy}{dx}+y\tan x=-\sin x\ln \left( \sec x \right)+\sin x-C\sin x+\sin x\ln \left( \sec x \right)+C\sin x$
Simplifying, we get
$\dfrac{dy}{dx}+y\tan x=\sin x$, which is the given differential equation.
Hence our answer is verified to be correct.