# Solving a simple differential equation

• Jan 12th 2010, 05:14 PM
xterminal01
Solving a simple differential equation
$\displaystyle 1st$

$\displaystyle y' = x/y^2$
$\displaystyle \int\text{ }y^2dy = \int\text{ }xdx$

$\displaystyle y^3/3 = x^2/2 + C$

$\displaystyle y^3 = 3x^2/2 + C$

$\displaystyle 2nd$

Find the particular solution that satisfies the initial condition.

$\displaystyle dr/ds$ = e^(r-s) when condition
r(0) = 1

$\displaystyle \int\text{ }1/e^r dr = \int\text{ } e^s ds$

$\displaystyle 1/e^r = e^s + C$

1/e^1= e^0+ C which means C= 1/e
• Jan 12th 2010, 05:19 PM
pickslides
Quote:

Originally Posted by xterminal01

Find the particular solution that satisfies the initial condition.

$\displaystyle dr/ds$ = e^(r-s) when condition
r(0) = 1

$\displaystyle \int\text{ }1/e^r dr = \int\text{ } e^s ds$

$\displaystyle 1/e^r = e^s + C$

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

$\displaystyle \frac{dr}{ds}= e^{r-s}= \frac{e^r}{e^s}$

gives

$\displaystyle \frac{dr}{e^r}= \frac{ds}{e^s}$

and

$\displaystyle \int e^{-r}~dr= \int e^{-s}~ds$

You got it from here?
• Jan 12th 2010, 05:37 PM
xterminal01
e^-r = e^-s + C ?
e^(-1) = e^(0) + c
1/e = C

Quote:

Originally Posted by pickslides
$\displaystyle \frac{dr}{ds}= e^{r-s}= \frac{e^r}{e^s}$

gives

$\displaystyle \frac{dr}{e^r}= \frac{ds}{e^s}$

and

$\displaystyle \int e^{-r}~dr= \int e^{-s}~ds$

You got it from here?

• Jan 13th 2010, 03:15 PM
xterminal01
Is this correct?
Quote:

Originally Posted by xterminal01
e^-r = e^-s + C ?
e^(-1) = e^(0) + c
1/e = C

• Jan 13th 2010, 03:26 PM
pickslides
From $\displaystyle \int e^{-r}~dr= \int e^{-s}~ds$

I get $\displaystyle -e^{-r}=-e^{-s}+C$

then

$\displaystyle e^{-r}=e^{-s}+C$

$\displaystyle -r=\ln(e^{-s}+C)$

using $\displaystyle r(0)=1$ to solve for $\displaystyle C$

$\displaystyle -1=\ln(e^{-0}+C)$

$\displaystyle e^{-1}=e^{-0}+C$

$\displaystyle e^{-1}=1+C$

$\displaystyle C=e^{-1}-1$
• Jan 13th 2010, 07:49 PM
xterminal01
Thanks, also Is the first eqq in my post done correctly as well?
Quote:

Originally Posted by pickslides
From $\displaystyle \int e^{-r}~dr= \int e^{-s}~ds$

I get $\displaystyle -e^{-r}=-e^{-s}+C$

then

$\displaystyle e^{-r}=e^{-s}+C$

$\displaystyle -r=\ln(e^{-s}+C)$

using $\displaystyle r(0)=1$ to solve for $\displaystyle C$

$\displaystyle -1=\ln(e^{-0}+C)$

$\displaystyle e^{-1}=e^{-0}+C$

$\displaystyle e^{-1}=1+C$

$\displaystyle C=e^{-1}-1$

• Jan 13th 2010, 07:52 PM
pickslides
Quote:

Originally Posted by xterminal01
$\displaystyle 1st$

$\displaystyle y' = x/y^2$
$\displaystyle \int\text{ }y^2dy = \int\text{ }xdx$

$\displaystyle y^3/3 = x^2/2 + C$

$\displaystyle y^3 = 3x^2/2 + C$

Finishing

$\displaystyle y^3 = \frac{3x^2}{2} + C$

$\displaystyle y = \sqrt[3]{\frac{3x^2}{2} + C}$