`(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`  on`ln|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)`