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

Expert Answers

An illustration of the letter 'A' in a speech bubbles

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`


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)`

See eNotes Ad-Free

Start your 48-hour free trial to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Get 48 Hours Free Access
Approved by eNotes Editorial Team