Question

solve dy/dx+y.tanx=secx​

Answer 1

$\textbf{Given equation is }$

$\frac{dy}{dx}+y\;tanx=secx$

$\text{This is linear differential equation in x}$

$\text{The solution of this equation is}$

$\boxed{\bf\,y\,e^{\int{P}\,dx}=\int\;Q\, e^{\int{P}\,dx}\,dx+c}$

$\text{Here P=tanx and Q=secx}$

$\text{Now,}$

$e^{\int{P}\,dx}$

$=e^{\int{tanx}\,dx}$

$=e^{log\,secx}$

$=secx$

$\text{The solution is}$

$\bf\,y\,e^{\int{P}\,dx}=\int\;Q\, e^{\int{P}\,dx}\,dx+c$

$y\,secx=\int\;secx\,secx \,dx+c$

$y\,secx=\int\;sec^2x \,dx+c$

$\implies\boxed{\bf\,y\,secx=tanx+c}$

$\therefore\textbf{The solution is y secx=tanx+c}$

Find more:

If y = y(x) is solution of the differential equation x. (dy/dx) + 2y = x²

satisfying y(1) = 1, then y(1/2)

is equal to (A) 7/64

(B) 1/4

(C) 49/16

(D) 13/16

[JEE Main 2019]

https://brainly.in/question/16030167