First order ode

• Sep 21st 2013, 07:05 PM
tylersmith7690
First order ode
For the following diferential equation

dy/dx = (2 co2x - sin2x + y2)/ 2 cos x , -pi/2<x<pi/2

show that the substitution y(x) = sin x + (1/u(x)) yields the deferential equation for u(x),

du/dx = -u tanx - (1/2) sec x

Hence fi nd the solution y(x) to the original deferential equation that satisfies the condition
y(0) = 2. Find the interval on which the solution to the initial value problem is defi ned.

Im having trouble doing this question. I have had about 4 attempts and can not seem to get to the du/dx they have. I'm solving the y= sinx +1/u to give dy/du= cosx -u-2
and then letting that = the original dy/dx. Is this the way to go about it or am i missing something? i also rearrange the substitution so u= 1/(y-sin x).
• Sep 21st 2013, 07:37 PM
junkwisch
Re: First order ode
I just realised that we have the exactly same question lol

http://mathhelpforum.com/calculus/22...-calculus.html

but I think our method is different, I'm working backward by integrate $\displaystyle \frac{dy}{dx}$ to give me the equation for y, then I also integrate $\displaystyle \frac{du}{dx}$ to get u
so i can sub them into $\displaystyle y(x)=sin(x)+\frac{-1}{u(x)}$