# Thread: First order Differential Equations

1. ## First order Differential Equations

Hi
I am having trouble with the following questions:
Solve the differential equations
1) $\frac{dy}{dx} = e^{x+y}$

$\frac{dy}{dx} = e^x * e^y$

$\frac{1}{e^y} * \frac{dy}{dx} = e^x$

not sure what i should do next

2) $\frac{dy}{dx} = 1+x-y-xy$

$y * \frac{dy}{dx}$

$y(1+x) * \frac{dy}{dx} = 1+x$

$y * \frac{dy}{dx} = \frac{1+x}{1+x}$

$y * \frac{dy}{dx} = 1$

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

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

P.S

2. $\displaystyle\frac{dy}{dx} = e^x \times e^y$

$\displaystyle\frac{dy}{dx}e^{-y} = e^x$

$\displaystyle e^{-y}~dy = e^x~dx$

$\displaystyle \int e^{-y}~dy = \int e^x~dx$

3. I'm not sure what you did with number 2, but it's linear and separable.

${dy\over dx}=1+x-y(1+x)=(1+x)(1-y)$

4. Originally Posted by matheagle
I'm not sure what you did with number 2, but it's linear and separable.

${dy\over dx}=1+x-y(1+x)=(1+x)(1-y)$
where does the +x go??

5. Matheagle has taken out $(1+x)$ as a factor. Its called factoring by grouping.

${dy\over dx}=1+x-y(1+x)=(1+x)-y(1+x)= (1+x)(1-y)$

${dy\over dx}= (1+x)(1-y)$

${dy\over (1-y)}= (1+x)~dx$

$\int {dy\over (1-y)}= \int(1+x)~dx$

6. This is my answers but it doesn't match the answer's in book

1)
$y-e^{-y} = e^x+C$
$e^{-y} = -(e^x+C)$
$y = -ln(-e^x-C)$

2)
$ln|1-y| = x+\frac{x^2}{2} + C$
$1-y = e^{x+\frac{x^2}{2} + C}$
$-y = -1+e^{x+\frac{x^2}{2} + C}$
$y=1-e^{x+\frac{x^2}{2} + C}$

7. Originally Posted by Paymemoney
This is my answers but it doesn't match the answer's in book

1)
$y-e^{-y} = e^x+C$
$e^{-y} = -(e^x+C)$
$y = -ln(-e^x-C)$

2)
$ln|1-y| = x+\frac{x^2}{2} + C$
$1-y = e^{x+\frac{x^2}{2} + C}$
$-y = -1+e^{x+\frac{x^2}{2} + C}$
$y=1-e^{x+\frac{x^2}{2} + C}$

Note:

-C can be 're-branded' as B,
e^C can be 're-branded' as A

since in both cases C is arbitrary and therefore B and A are also arbitrary.

1) actually got it right $y=-ln(k-e^x)$ where k=-c

2) $y=1+ke^{-(x+\frac{x^2}{2})}$

9. Originally Posted by Paymemoney
[snip]
2)
$ln|1-y| = x+\frac{x^2}{2} + C$ Mr F says: There should be a negative in front of the ln. Make the necessary corrections in your working below and then recall from my previous post the note about 're-branding' C ....

$1-y = e^{x+\frac{x^2}{2} + C}$
$-y = -1+e^{x+\frac{x^2}{2} + C}$
$y=1-e^{x+\frac{x^2}{2} + C}$
..

10. i don't understand why should there be a negative at the front of ln?

The intergal of $\frac{1}{1-y}$ $= ln|1-y|$

11. $\displaystyle{\int\frac{dy}{1-y}=-\int\frac{dy}{y-1}=-\ln|y-1|+C.}$

The rule you're thinking of is that

$\displaystyle{\int\frac{dy}{y-a}=\ln|y-a|+C.}$
However, the sign of $y$ must be positive for that rule to work.

12. Originally Posted by Ackbeet
$\displaystyle{\int\frac{dy}{1-y}=-\int\frac{dy}{y-1}=-\ln|y-1|+C.}$

The rule you're thinking of is that

$\displaystyle{\int\frac{dy}{y-a}=\ln|y-a|+C.}$
However, the sign of $y$ must be positive for that rule to work.
oh i see now