D Y D X − Y Tan X = E X Sec X - Mathematics

Sum

$\frac{dy}{dx} - y \tan x = e^x \sec x$

Solution

We have,

$\frac{dy}{dx} - y \tan x = e^x \sec x$

$\text{Comparing with }\frac{dy}{dx} + Py = Q,\text{ we get}$

$P = - \tan x$

$Q = e^x \sec x$

Now,

$I . F . = e^{\int - \tan x\ dx}$

$= e^{- \log\left| \left( \sec x \right) \right|}$

$= e^{\log\left| \left( \cos x \right) \right|}$

$= \cos x$

So, the solution is given by

$y \cos\ x = \int\left( \cos x\ e^x \sec x \right) dx + C$

$\Rightarrow y \cos\ x = \int e^x dx + C$

$\therefore y \cos\ x = e^x + C$

Is there an error in this question or solution?

APPEARS IN

RD Sharma Class 12 Maths
Chapter 22 Differential Equations
Revision Exercise | Q 42 | Page 146