Thread: FInding particular solutions of differential equations

1. FInding particular solutions of differential equations

I got stuck on these on my homework, and was hoping someone could help be out.

1. ysqrt(1-x^2)y' - xsqrt(1-y^2) = 0
Initial condition is y(0) = 1

2. dr/ds = e^(r+s)
Initial condition is r(1) = 0

Thanks!

I got stuck on these on my homework, and was hoping someone could help be out.

1. ysqrt(1-x^2)y' - xsqrt(1-y^2) = 0 Mr F says: $y \, \sqrt{1 - x^2} \, \frac{dy}{dx} = x \sqrt{1 - y^2} \Rightarrow \frac{y}{\sqrt{1 - y^2}} \, dy = \frac{x}{\sqrt{1 - x^2}} \, dx$ ......

Initial condition is y(0) = 1

2. dr/ds = e^(r+s) Mr F says: Writing $e^{r + s} = e^r e^s$ might make the DE more obviously seperable to you .....

Initial condition is r(1) = 0

Thanks!
..

I got stuck on these on my homework, and was hoping someone could help be out.

1. ysqrt(1-x^2)y' - xsqrt(1-y^2) = 0
Initial condition is y(0) = 1

2. dr/ds = e^(r+s)
Initial condition is r(1) = 0

Thanks!

The first is seperable...

$\frac{y}{\sqrt{1-y^2}}dy=\frac{x}{\sqrt{1-x^2}}dx$

Just integrate both sides

The second is also seperable.

note that

$e^{r+s}=e^re^s$

This gives

$e^{-r}dr=e^sds$

take the integral of both sides