Answer:
The answer is "[tex]u = e^{\frac{9}{2}}-\frac{1}{2} \cos v^2\\[/tex]"
Step-by-step explanation:
In the given-question some of the equation is missing, which is defined as follows:
[tex]\bold{\frac{du}{dv} = u v \sin v^2} \\\\ \bold{u(0) = e^4}[/tex]
Solution:
[tex]\to \frac{du}{dv} = u v \sin v^2 \\\\ \to \frac{du}{u} = v \sin v^2 \ dv\\\\ \to \int \frac{du}{u} = \int v \sin v^2 \ dv +c\\\\[/tex]
[tex]|n|u| = -\frac{1}{2} \cos v^2 +c\\\\\ given u(0) = e^4 \\\\\to |n (e^4) = -\frac{1}{2} \ cos (0) +c\\\\\to 4= -\frac{1}{2} +c\\\\\to c= 4+\frac{1}{2}\\\\\to c= \frac{8+1}{2}\\\\\to c= \frac{9}{2}\\\\[/tex]
[tex]\to |n|u| =-\frac{1}{2} \cos v^2+\frac{9}{2}\\\\\\\boxed{u = e^{\frac{9}{2}}-\frac{1}{2} \cos v^2}\\[/tex]
