# Solving a simple homogeneous linear DE

1. Apr 17, 2012

### Nano-Passion

I feel that it may be redundant to rewrite the whole problem. I just need to know how the book got from point to point b.

The book says that
$$e^{-3x} \frac{dy}{dx} - 3y (e^{-3x}) = 0$$
is the same as $$\frac{d}{dx}(e^{-3x}y) = 0$$

How? I tried dividing and multiplying by some variables to get the same expression but all was in vain.

Last edited: Apr 17, 2012
2. Apr 17, 2012

### d2j2003

product rule for differentiation...

3. Apr 17, 2012

### Nano-Passion

Have patience with me because I'm still in Calculus II so I tend to overlook things that may appear simple to you. Anyways, I don't get the same answer for some reason.

$$e^{-3x} \frac{dy}{dx} = 3y (e^{-3x}$$
Lets call the left side eq (1) & the right side eq (2).
Derivative of eq (1): let u = e^(-3x) & v = dy/dx

What is the derivative of dy/dx? This seems to be the reason why I'm getting the wrong answer. I first thought d^2y/dx^2 but plugging that in ultimately leaves me with an incorrect expression.

4. Apr 17, 2012

### Staff: Mentor

That is irrelevant to the problem and is not the question to ask. The expression comes from differentiating y with respect to x.
The original DE is e-3x*dy/dx - 3ye-3x = 0.

The idea is to recognize that the left side is the same as (i.e., equal to) d/dx(ye-3x). To convince yourself that this is true, carry out the differentiation of ye-3x, differentiating y implicitly.

5. Apr 17, 2012

### d2j2003

d/dx [f(x)g(x)] = f'(x)g(x)+f(x)g'(x) (product rule) does this look familiar?

6. Apr 17, 2012

### Nano-Passion

Thanks it makes sense actually. But I couldn't derive it algebraically.

I couldn't differentiate it implicitly. I was doing a bit of research to refresh my memory but it doesn't seem that you can use implicit differentiation for an expression. The implicit differentiation in calculus textbooks is reserved for the solving of equations and not expressions.

Yes I know the product rule, it is what I attempted in my previous post. d/dx uv=u'v+v'u

7. Apr 17, 2012

### Staff: Mentor

Your understanding is flawed. You can certainly differentiate an expression. For example, assuming that y represents a differentiable function of x, d/dx(x2y) = 2xy + x2dy/dx.

8. Apr 17, 2012

### Nano-Passion

I did some snooping around the textbook and I realized what has been slowing me down soo much.

Recall that I'm in Calculus II at the moment. So I haven't studied the chain rule for the function of two variables. That is what is likely slowing me down.

Edit:

When you said differentiate it implicitly, I thought you meant to use the procedure called implicit differentiation supplied in Calculus I.

Last edited: Apr 17, 2012
9. Apr 17, 2012

### Nano-Passion

But like I said before, differentiating it gives me an incorrect expression. I'm probably doing something wrong. Here is the differentiation for eq. (2).

$$\frac{d}{dx} 3ye^{-3x}$$
Using the product rule with the chain rule: $$= 3 (\frac{dy}{dx} e^{-3x} + -3e^{-3x}y)$$

10. Apr 17, 2012

### sharks

Do you know how to find the Integrating Factor?

There are actually a few steps not shown in your book going from the first line to the second one.

In order to solve a linear homogeneous ODE, you need to first write it in the correct form.
$$e^{-3x} \frac{dy}{dx} - 3y (e^{-3x}) = 0$$
To convert the linear ODE in standardized form, you must first divide by $e^{-3x}$, which gives:
$$\frac{dy}{dx} - 3y = 0$$
From there, you need to find the integrating factor, which is $e^{-3x}$

Then, you just follow the usual steps:
$$\frac{d(μy)}{dx} = μQ$$
Therefore,
$$\frac{d(e^{-3x}y)}{dx} = 0$$

Now, to continue the procedure in solving this ODE, you would integrate both sides and then finally, divide throughout by the integrating factor, to get y. Don't forget to include your arbitrary constant of integration in the final answer.

Check if this is the answer:
$$y = \frac{A}{e^{-3x}}$$
where A is the constant of integration.

Last edited: Apr 17, 2012
11. Apr 17, 2012

### Staff: Mentor

No, not quite.

It should be
$$\frac{d}{dx} ye^{-3x}$$

Try again. That extra 3 you had is throwing you off.

12. Apr 17, 2012

### Staff: Mentor

I'm guessing that the OP does not know this technique, which in any case is not needed.
No, there are not. The first line in post #1 can be obtained by differentiating both sides of the equation ye-3x = 0.
The equation is already in a form that can be used to solve it. No integration factor is needed.

13. Apr 18, 2012

### Nano-Passion

I checked and rechecked but I don't see any extra 3?

Edit

Nvm, rechecking my calculations.

Last edited: Apr 18, 2012
14. Apr 18, 2012

### Nano-Passion

There was a miscommunication, I misinterpretated what you meant by the left side because of the way I wrote the equation prior to that.

Thanks, I understand everything now.

15. Apr 18, 2012

### d2j2003

ok so $\frac{d}{d}(e^{-3x})(y) = (by product rule) [\frac{d}{dx}(e^{-3x})]y + e^{-3x}[\frac{d}{dx}(y)] = (by chain rule) e^{-3x}[\frac{d}{dx}(-3x)]y+e^{-3x}[\frac{d}{dx}(y)]$

then go from there..