(du)/(dv) = uvsin(v^2) , u(0) = 1 Find the particular solution that satisfies the initial condition

An ordinary differential equation (ODE) has differential equation for a function with single variable. A first order ODE follows  .
In the given problem: (du)/(dv)=uvsin(v^2) ,  we may apply variable separable differential equation in a form of   .
Divide both sides by "u" and cross-multiply dv  to set it up as:
(du)/u=vsin(v^2) dv.
Apply direct integration: int(du)/u=int vsin(v^2) dv.
For the left sign, we follow the basic integration formula for logarithm:
int (du)/u = ln|u|
For the right side, we follow the basic integration formula for sine function:
Let: w=v^2 then dw = 2v*dv or (dw)/2 =v dv .
The integral becomes:
intvsin(v^2) dv= intsin(v^2) * vdv
                       =intsin(w) *(dw)/2
                      = (1/2) int sin(w) dw
                      = (1/2)*(-cos(w))+C
                      =-cos(w)/2+C
Plug-in w=v^2 on -cos(w)/2+C , we get:
intvsin(v^2) dv=-cos(v^2)/2+C
Combing the results, we get the general solution of differential equation as:
ln|u| = -cos(v^2)/2+C
 
To solve for the arbitrary constant (C) , apply the initial condition u(0)=1  onln|u| = -cos(v^2)/2+C :
ln|1| = -cos(0^2)/2+C
0 = -1/2+C
C = 0+1/2
C=1/2
Plug-in  C= 1/2 in ln|u| = -cos(v^2)/2+C , we get 
ln|u| = -cos(v^2)/2+1/2
 u = e^(-cos(v^2)/2+1/2)