# First order ODE

1. Jul 12, 2005

### Benny

Q. By finding a suitable integrating factor, solve the following equation:

$$\left( {x + y^3 } \right)y' = y$$ (treat y as the independent variable).

Answer: Exact equation is $$y^{ - 1} \left( {\frac{{dx}}{{dy}}} \right) - xy^{ - 2} = y$$ leading to $$x = y\left( {k + \frac{{y^2 }}{2}} \right)$$.

I'm having trouble indentifying the type of the DE.

$$\left( {x + y^3 } \right)y' = y$$

$$ydx - \left( {x + y^3 } \right)dy = 0$$

Writing the equation as I did above makes it kind of look like an exact(or one that can be made exact) DE but I'm still not sure what to do. I'm also being unsure about how to "treat y as the independent variable." All I can gather from that instruction is to find x = x(y). Can someone please help me out?

2. Jul 12, 2005

### Pyrrhus

You will need an integrating factor to solve this. It's an inexact DE. When they say "treat y as the independent variable", probably means look for a integrating factor in function of y $\mu = f(y)$.

3. Jul 12, 2005

### Benny

Thanks for the help Cyclovenom.

4. Jul 13, 2005

### mathelord

express the equation in the form of mdx+ndx=0.then u get the integrating factor by raising to te power of e the integral of (1/n)(dm/dy - dn/dx).THEN YOU GET AN EXACT EQUATION WHICH IS SEPARABLE

5. Jul 13, 2005

### GCT

you'll need to treat x as the dependent variable

$$\frac{dy}{dx}= \frac{y}{x+y^{3}}$$

$$\frac{dx}{dy}= \frac{x+y^{3}}{y}$$

$$\frac{dx}{dy} - \frac{x}{y} = y^{2}$$

$$u(y)=e^{ \int \frac{1}{y} ~ dy}$$

I'm sure you can go on from here

mathelord, can you elaborate more on your approach (preferably through latex)

6. Jul 13, 2005

### Benny

Yeah thanks for the help I solved it when I got up. I think that in using the method that I did, I ignored the instruction to use the given method. Anyway after seeing GCT's answer it's become clear what I needed to do.

I'm pretty sure mathelord is referring to the technique where you get a solution of the form F(x,y) = c(in this case you can solve for x in terms of y). The dn/dx, dm/dy should be partial derivatives I'd say. Also I don't think it's possible to get an integrating factor in terms of x or y only, by using the equation that mathelord has given. I used $$\mu \left( y \right) = \exp \left( {\int {\frac{{\frac{{\partial N}}{{\partial x}} - \frac{{\partial M}}{{\partial y}}}}{M}} dy} \right)$$ where the original DE has been written in the form $$M\left( {x,y} \right)dx + N\left( {x,y} \right)dy = 0$$.

I have another question that I would like some help with.

$$\frac{{dy}}{{dx}} - \frac{{y^2 }}{{x^2 }} = \frac{1}{4}$$

The general solution I get is:

$$y = \frac{x}{2}\left( {\frac{2}{{ - 2c - 2\log \left| x \right|}} + 1} \right)$$

I am also given two ICs y(1) = 1 and y(1) = 1/2. For y(1) = 1 I get $$y = \frac{x}{{2 - \log \left| x \right|}} + \frac{x}{2}$$ which is the same as the book's answer. However, for y(1) = 1/2 the book's answer is "singular solution y = x/2." Can someone explain what that means to me? Using y(1) = 1/2 I get an answer similar to the one I got for y(1) = 1 so I don't know where y = x/2 came from.

Last edited: Jul 13, 2005
7. Jul 13, 2005

### HallsofIvy

Staff Emeritus
Interesting! There were actually three different suggestions each of which gives the correct integrating factor!

8. Jul 13, 2005

### saltydog

What's the answer then? How do you know it's correct? It's implicitly defined. Can you generate a plot from it? Say, for the initial conditions y(0)=1? Just suggestions that's all.

9. Jul 13, 2005

### Hurkyl

Staff Emeritus
How? Doesn't it require you to solve: &nbsp; 1 - 1/c = 1 &nbsp; for c?

If you go back through your work, I guarantee you'll find you made an unwarranted assumption somewhere along the way.

(For those who like sloppy reasoning, you could get the "singular" solution by setting c = &infin;)

10. Jul 13, 2005

### saltydog

I know. I'm a pain. Here's the plot:

#### Attached Files:

• ###### implicitplot.JPG
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11. Jul 13, 2005

### Benny

Hurkyl - I must have made a mistake while I was solving the second one. Seeing as I had a general solution I didn't bother writing the second equation down on paper which might've been what caused me to overlook the equation you presented.

Saltydog - I was referring to my first question when I said I was able to solve it. I assume that the answer I got is correct since it agrees with the book's answer. I don't really have a habit of carrying out the differentiation and substituting back into every DE I try to solve.