1. First order Differential Equations

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

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

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

not sure what i should do next

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

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

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

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

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

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

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

P.S

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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 $\displaystyle y=-ln(k-e^x)$ where k=-c

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

9. Originally Posted by Paymemoney
[snip]
2)
$\displaystyle ln|1-y| = x+\frac{x^2}{2} + C$ 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 ....

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

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

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

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

The rule you're thinking of is that

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

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

The rule you're thinking of is that

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