Question

For the given differential equation, find the general solution: $\frac{dy}{dx} + (sec\ x) y= tan\ x\biggr\lgroup 0\leq x\leq \frac{\pi}{2} \biggr\rgroup$

Answer 1

Answer:

Step-by-step explanation:

Concept:

The solution of

$\frac{dy}{dx}+Py=Q\: is\\\\y.e^{\int{P}dx}=\int{Q.e^{\int{P}dx}dx}+c$

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

comparing this with

$\frac{dy}{dx}+Py=Q$

we get

P=secx

Q=tanx

Integrating factor

$=e^{\int{P}dx}\\\\=e^{\int{secx}\:dx}\\\\=e^{log(secx+tanx)}\\\\=secx+tanx$

The solution is

$y.e^{\int{P}dx}=\int{Q.e^{\int{P}dx}dx}+c\\\\y.(secx+tanx)=\int{tanx(secx+tanx)dx}+c\\\\y.(secx+tanx)=\int{[secx.tanx+tan^2x]dx}+c\\\\y.(secx+tanx)=\int{secx.tanx}\:dx+\int{tan^2x}\:dx+c\\\\y.(secx+tanx)=\int{secx.tanx}\:dx+\int{(sec^2x-1)}\:dx+c\\\\y.(secx+tanx)=secx+tanx-x+c$