# Help with Separable Differential Equations

1. Sep 30, 2008

### wanamaa

Hi there, I was working on two of my homework problems for calculus and I'm stuck.

First equation is: dy/dx=e^(x+y), solving for y
So far here is what I have:
dy/dx=(e^x)(e^y)
therefore dy/dx=(e^x)/(1/e^y)
INT(1/e^y)dy=INT(e^x)dx
from the integration, my calculator comes up with -e^-y=e^x, and now I'm stuck

second equation is: dy/dx=x/(1+2y)
So far here is what I have:
INT(1+2y)dy=INT(x)dx
from that, I get y^2+y=(x^2)/x + C
which leads to y(y+1)=(x^2)/x + C
now I'm stuck here too.

Please help, I feel bad posting a question as my first post, but any help would be awesome. This assignment isn't due for a week so I'm kinda in a hurry but not. Thanks guys!

2. Sep 30, 2008

$$-e^{-y} = e^x + C$$

where $$C$$ is a constant. You need to isolate $$y$$
How would you solve this equation:

$$-e^{-y} = 8$$

The same type of steps will work for you in current setting.

For the second question, you get to

$$\int (1+2y) \, dy = \int x \, dx$$

Regarding the integral on the right, why do you write

$$\int x \, dx = \frac{x^2}{x}$$

For the integral on the left: rather than integrating the two terms alone, treat $$1 + 2y$$ as a single expression and find an indefinite integral for it (use a $$u [tex]-substitution if it helps you see the result) 3. Sep 30, 2008 ### wanamaa I wrote that because that's what the integral for x dx is. Unless my TI-89 is lying to me.... I understand where I forgot the C, but with what you wrote, where did the 8 come from? Is it just some arbitrary number to help solve for y or does it have any significance? Maybe I completely missed something here. 4. Sep 30, 2008 ### wanamaa whoops, looks like it wouldn't let me copy the equations you wrote up above. Weird. 5. Sep 30, 2008 ### statdad First: if your calculator is really telling you that [tex] \int x \, dx = \frac{x^2}{x}$$

it does have a problem with integrity

Hint:

$$\frac{d}{dx}\left(\frac{x^2}{2}\right) = \frac{2x}{2} = x$$

Second, my equation

$$-e^{-y} = 8$$

doesn't come from your problem - I merely asked how you would go about solving it for $$y$$. If you can solve the problem I posed (and posed here again), then the
same steps will solve the one you need to solve.

6. Sep 30, 2008

### wanamaa

oh snap! I screwed up writing the notation for x^2/x. I meant x^2/2. Sorry, now I feel stupid .

To solve the second equation, I am given an initial value of y(-1)=0. Forgot to mention that in the very beginning oops. I just don't know if I need to refine the equation I found near the end or if I need to restart that problem and work it through another method.

7. Sep 30, 2008

"Sorry, now I feel stupid ."

Don't - proofreading one's own work when you are writing it is difficult enough; proofreading when you are typing on a computer is more difficult.

If you've done the integrations correctly, and included the constant $$C$$ in your indefinite integrals, then you can incorporate the initial condition from those equations. (you have a formula for the function $$y$$, just evaluate each side at $$x = -1$$ to determine the appropriate value for the constant of $$C$$.

8. Sep 30, 2008

### HallsofIvy

Staff Emeritus
Do you find it at all embarrasing that you used a calculator to integrate e-y dy and exdx?

If you had done it by hand, you might have recognized that the correct answer would be
-e-y= ex+ C where C is the "constant of integration. Now, why are you stuck? That is, what are you trying to do? Solve for y? My first suggestion would be to multiply the entire equation by -1 so e-y= -ex+ C and the "get rid of" the exponential by using its inverse function, ln: take the logarithm of both sides of the equation.

Again, what is it you are trying to do? (But congratulations on remembering the C!) Solve for y? Are you actually required to do that?
What you have is a quadratic equation for y isn't it- y2+ y- (x2/2+ C)= 0. If you must, use the quadratic formula with a= 1, b= 1, c= -(x2/2+ C).

9. Sep 30, 2008

### wanamaa

Basically I am trying to solve for the separable differential equation for y, and I also was given an initial y value (which I forgot in my first post but posted it I think in my 3 or 4th post on this thread), which is y(-1)=0 for both equations. I know how to input initial values, I just got to those two points on each problem and am stuck algebraically where to go, because I can't take the log of -e^-y, it's a non-real answer.

And yes, I do feel stupid checking on my calculator for the integrals, I'm always unsure of myself when it comes to math

10. Sep 30, 2008

### wanamaa

Yay, I was close on that one! Ok, that is what I was trying to do in the beginning and kept running into the "can't take a log of a negative number" problem. But won't I run into the same problem with the newly made "-e^x - C" or would I just rearrange the variables, like "-C-e^x"? Sorry for th stupid questions

Last edited: Sep 30, 2008