`(dr)/(ds) = e^(r-2s) , r(0)=0` Find the particular solution that satisfies the given initial condition.

Expert Answers

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

This differential equation can be solved by separating the variables.

`(dr)/(ds) = e^(r - 2s)`

Dividing by e^r and multiplying by ds results in the variables r and s on the different sides of the equation:

`(dr)/e^r = e^(-2s)ds`

This is equivalent to

`e^(-r) dr = e^(-2s)ds`

Now we can take the integral of the both sides of the equation:

`-e^(-r) = 1/(-2)e^(-2s) + C` , where C is an arbitrary constant.

From here, `e^(-r) = 1/2e^(-2s) - C`

and `-r = ln(1/2e^(-2s) - C)`

or `r = -ln(1/2e^(-2s) - C)`

Since the initial condition is r(0) = 0, we can find the constant C:

`r(0) = -ln(1/2e^(-2*0) - C) = -ln(1/2 - C) = 0`

This means `1/2 - C = 1`

and `C = -1/2`

Plugging C in in the equation for r(s) above, we can get the particular solution:

`r = -ln((e^(-2s) + 1)/2)` . This is algebraically equivalent to

`r = ln(2/(e^(-2s) + 1))` . This is the answer.

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