differential equation

• Mar 13th 2013, 10:52 AM
vervas
differential equation
hi guys,
I´m struggling with this question on differential equations: dy/dx = (1+siny)/cosy.
I separated x and y, integrated and got to the point ln (1+siny)=x+const. but can´t find any way to express this equation in the form y= something.
thanks a lot for any help!
• Mar 13th 2013, 11:14 AM
Soroban
Re: differential equation
Hello, vervas!

Quote:

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

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

But can't solve for $y.$

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

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

. . . . $\sin y \;=\;Ce^x-1 \quad\Rightarrow\quad y \;=\;\sin^{-1}\!\left(Ce^x-1\right)$
• Mar 13th 2013, 11:25 AM
HallsofIvy
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 $2\pi$ will give other solutions.
• Mar 13th 2013, 11:25 AM
vervas
Re: differential equation
thank you! I got it except this part: e^x*e^c = C* e^x, could you explain that please?
• Mar 13th 2013, 11:38 AM
vervas
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 it´s to most useful expression