Question

solve the differential equation dy/dx+ytanx=y^2secx

Answer 1

Please see the attachment

Answer 2

Answer:

The solution is $\frac{\cos x}{x+C}$

Step-by-step explanation:

Given: equation

To find: a solution for the equation

Solution:

$\frac{d y}{d x}+y \tan x=y^{2} \sec x$

Dividing both sides by $y^{2}$, we get

$\frac{1}{y^{2}} \frac{d y}{d x}+\frac{1}{y} \tan x=\sec x$

Putting $\frac{1}{y}=u$

$\Rightarrow-\frac{1}{y^{2}} \frac{d y}{d x}=\frac{d u}{d x}$

So, from (1), we get

$\begin{aligned}&-\frac{d u}{d x}+(\tan x) u=\sec x \\&\Rightarrow \frac{d u}{d x}+(-\tan x) u=-\sec x\end{aligned}$

which is linear differential equation of the form $\frac{d y}{d x}+p(x) y=q(x)$

whose solution is given by $y \mu(x)=\int q(x) \mu(x) d x+C ,$

where $\mu(x)=e^{\int p(x) d x}$ is the integrating factor.

Here, $p(x)=-\tan x$

$\text { So, } \mu(x)=e^{\int-\tan x d x}=\cos x$

The solution is given by

$\begin{aligned}&u \cos x=\int \cos x \sec x d x+C \\&\Rightarrow u \cos x=x+C \\&\Rightarrow \frac{\cos x}{y}=x+C \\&\Rightarrow y=\frac{\cos x}{x+C}\end{aligned}$