# Math Help - Solving a simple differential equation

1. ## Solving a simple differential equation

$1st$

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

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

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

$2nd$

Find the particular solution that satisfies the initial condition.

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

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

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

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

2. Originally Posted by xterminal01

Find the particular solution that satisfies the initial condition.

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

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

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

1/e^1= e^0+ C which means C= 1/e
$\frac{dr}{ds}= e^{r-s}= \frac{e^r}{e^s}$

gives

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

and

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

You got it from here?

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

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

gives

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

and

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

You got it from here?

4. Is this correct?
Originally Posted by xterminal01
e^-r = e^-s + C ?
e^(-1) = e^(0) + c
1/e = C

5. From $\int e^{-r}~dr= \int e^{-s}~ds$

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

then

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

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

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

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

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

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

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

6. Thanks, also Is the first eqq in my post done correctly as well?
Originally Posted by pickslides
From $\int e^{-r}~dr= \int e^{-s}~ds$

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

then

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

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

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

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

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

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

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

7. Originally Posted by xterminal01
$1st$

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

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

$y^3 = 3x^2/2 + C$
Finishing

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

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