 #1
 143
 0
For the following problem
[itex]\frac{dw}{d\theta}=\theta w^{2}sin(\theta^{2}), w(0)=1[/itex]
I am not able to obtain the solution
[itex]w=\frac{2}{1+cos(\theta^{2})}[/itex]
Can anyone point out my mistake?
I have attached my working out in a picture format below (may need to enlarge it)
thanks
[itex]\frac{dw}{d\theta}=\theta w^{2}sin(\theta^{2}), w(0)=1[/itex]
I am not able to obtain the solution
[itex]w=\frac{2}{1+cos(\theta^{2})}[/itex]
Can anyone point out my mistake?
I have attached my working out in a picture format below (may need to enlarge it)
thanks
Attachments

34 KB Views: 366

28.6 KB Views: 371