Re: differential equation

Hello, vervas!

Quote:

$\displaystyle \frac{dy}{dx} \:=\: \frac{1+\sin y}{\cos y}$

I got: .$\displaystyle \ln(1+\sin y)\:=\:x+c$

But can't solve for $\displaystyle y.$

We have: .$\displaystyle \ln(1+\sin y) \:=\:x + c$

Exponentiate: .$\displaystyle 1 + \sin y \:=\:e^{x+c} \:=\:e^x\cdot e^c \:=\:Ce^x$

. . . . $\displaystyle \sin y \;=\;Ce^x-1 \quad\Rightarrow\quad y \;=\;\sin^{-1}\!\left(Ce^x-1\right)$

Re: differential equation

Are you **required** to solve for y? Many first order differential equations are satisified by some f(x, y)= constant where y is NOT a function of x. Soroban's solution (using the principle arc cosine) gives one solution out of many. Adding multiples of $\displaystyle 2\pi$ will give other solutions.

Re: differential equation

thank you! I got it except this part: e^x*e^c = C* e^x, could you explain that please?

Re: differential equation

to HallsofIvy

well, the question says : find the general solution of following equations, so it seems like I am not required to express it in the form y = ... but I though itīs to most useful expression