Thread: Solving a simple differential equation

Find the particular solution that satisfies the initial condition.

= e^(r-s) when condition r(0) = 1





1/e^1= e^0+ C which means C= 1/e

Originally Posted by xterminal01

Find the particular solution that satisfies the initial condition.

= e^(r-s) when condition r(0) = 1





1/e^1= e^0+ C which means C= 1/e

gives



and



You got it from here?

e^-r = e^-s + C ?
e^(-1) = e^(0) + c
1/e = C

Originally Posted by pickslides

gives



and



You got it from here?

Is this correct?

Originally Posted by xterminal01

e^-r = e^-s + C ?
e^(-1) = e^(0) + c
1/e = C

From

I get

then





using to solve for

Thanks, also Is the first eqq in my post done correctly as well?

Originally Posted by pickslides

From

I get

then





using to solve for

Originally Posted by xterminal01

Finishing