Question

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

Answer 1

Hint: The given differential equation is a first degree linear differential equation of the form $\dfrac{dy}{dx}+Py=Q$. The solution is given by:
$y\times I.F=\int{Q\times I.F.dx+c}$
where $I.F={{e}^{\int{Pdx}}}$
So we will first calculate an integrating factor known as I.F then we will substitute it in the equation of solution.

Complete step-by-step answer:
We have been given the differential equation $\dfrac{dy}{dx}+y\tan x=\cos x$.
It is of the form $\dfrac{dy}{dx}+Py=Q$ known as first degree differential equation and the solution is given by,
$y\times I.F=\int{Q\times I.Fdx+c}$
where $I.F={{e}^{\int{Pdx}}}$
So we have where $P=\tan x$ and $Q=\cos x$.
Now we will calculate the value of I.F as follows:
$I.F={{e}^{\int{Pdx}}}={{e}^{\int{\tan xdx}}}$
Since we know that $\int{\tan xdx}=ln\left( \sec x \right)$
$\Rightarrow I.F={{e}^{ln\left( \sec x \right)}}$
Also, we know that ${{e}^{lnt}}=t$
$\Rightarrow I.F=\sec x$
So the equation of the given differential equation is given as follows:
$y\sec x=\int{\sec x\cos xdx+c}$
Since we know that $\sec x=\dfrac{1}{\cos x}$

& \Rightarrow y\sec x=\int{\dfrac{1}{\cos x}\times \operatorname{cosxdx}}+c \\\ & \Rightarrow y\sec x=\int{dx+c} \\\ \end{aligned}$$ $$\Rightarrow y\sec x=x+c$$, where c is any arbitrary constant. Therefore, the solution of the given differential equation is equal to $$y\sec x=x+c$$. Note: Be careful while finding the value of I.F, i.e. integrating factor. Sometimes we just forget that the term $$\int{Pdx}$$ is on the power to ‘e’ and we use the I.F is equal to the value obtained by $$\int{Pdx}$$ which gives us the wrong result. Also, don’t forget to mention plus ‘c’ in the final solution which is an arbitrary constant.