# Differential equation.

1. Nov 20, 2013

### Mutaja

1. The problem statement, all variables and given/known data

1st problem - is this correctly done?

$\frac{dy}{dx}$ = ($x^2$ - 1) $y^2$ , y(0) = 1

2nd problem - I really need help with this one.

xy' - y = $3x^2$ , y(1) = 1

3. The attempt at a solution

1st problem:

$\frac{dy}{dx}$ = ($x^2$ - 1) $y^2$ , y(0) = 1

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

$y^{-2}$ dy = ($x^2$ -1) dx

∫$y^{-2}$ dy = ($x^2$ - 1) dx

$-y^{-1}$ = ($\frac{x^3}{3}$ - x) +c

$-1^{-1}$ = c

c = -1

$-y^{-1}$ = ($\frac{x^3}{3}$ - x) - 1.

2nd problem:

xy' - y = $3x^2$ , y(1) = 1

-y + xy' = $3x^2$

My = -1, Nx = 1

μ(x)(-y) + μ(x)(xy') = μ(x) $3x^2$

Am I onto something here? Any help or guidelines is highly appreciated.

2. Nov 20, 2013

### Saitama

Your attempt for 1st is correct.

For the second one, observe that the given differential equation can be converted to a linear D.E.

3. Nov 20, 2013

### phyzguy

For the second one, try to write the left-hand side as the derivative of something.

4. Nov 20, 2013

### Staff: Mentor

For the first one, you can always check that your solution is correct, which is a good habit to form. If your solution satisifies the initial condition and the differential equation, then you're golden.

5. Nov 20, 2013

### Saitama

Very clever! :)

6. Nov 20, 2013

### Mutaja

Thanks for your input! From the beginning I assume, not from the 'end of my progress'?

Edit: Either way, I'm not sure what to write...

Last edited: Nov 20, 2013
7. Nov 20, 2013

### Staff: Mentor

From the beginning. If you stare at xy' + y long enough, you might recognize it as the derivative (with respect to x) of something.

Edit: It should have been xy' - y, not xy' + y.

Then you can write d/dx(<something>) = 3x2. If you know that the derivative of something is 3x2, then that something is equal to what?

Last edited: Nov 20, 2013
8. Nov 20, 2013

### Mutaja

I'm not sure what to do, but am I onto something here?

$\frac{d}{dx}$ x y' - y = $3x^2$

y'-y = $3x^2$

I'm slightly confused.

Thanks a lot for your help, even though I don't understand what to do yet, I appreciate it.

9. Nov 20, 2013

### phyzguy

This isn't right.

$$\frac{d}{dx} (x y') \neq y'$$
$$\frac{d}{dx} (x y') = y' + x y''$$

Remember: $$\frac{d}{dx} (f g) =\frac{df}{dx} g + f \frac{dg}{dx}$$

Keep thinking along these lines.

10. Nov 20, 2013

### ehild

You can rearrange the equation as

$\frac{xy' - y}{x^2} = 3$

Do you recognize the left hand side as the derivative of a fraction?

ehild

(y/x)'

11. Nov 20, 2013

### Mutaja

I can see and understand what you've done - just divided by $x^2$. I do not, however, understand how I'm supposed to see that $\frac{xy' - y}{x^2}$ = (y/x)'.

Using $$\frac{d}{dx} (f g) =\frac{df}{dx} g + f \frac{dg}{dx}$$

I get that $$\frac{d}{dx} (x y') =\frac{dx}{dx} y' + x \frac{dy'}{dx}$$

$$\frac{d}{dx} (x y') = y' + x \frac{dy'}{dx}$$

And I'm stuck and don't understand what I'm supposed to do. These are methods I haven't yet seen in our lectures thus I don't have them in my notes.

I'm sorry I don't understand what you're trying to tell/learn me, but I'm trying my best.

Thanks a lot for helping me, I really appreciate it.

12. Nov 20, 2013

### Staff: Mentor

You're going the wrong way with it. We're not suggesting that you find the derivative of something, but rather guess what something should be to get a particular expression as its derivative.

For example, this is NOT what you're doing : d/dx(<given>) = <??>. Instead, it's figure out what <??> needs to be in this equation: d/dx(<??>) = <given>.

You can ignore my hint. I misread a sign in the problem. ehild's hint is more appropriate.

13. Nov 20, 2013

### ehild

You certainly have seen what is the derivative of f/g.

$(f/g) ' = \frac{f ' g - f g '}{g^2}$

f=y(x), g=x in this problem.

ehild

14. Nov 20, 2013

### Mutaja

So I'm not supposed to 'solve' the left hand side to get (y/x)' - I'm supposed to see it? Your example here confuses me even more.

Is there any chance you can tell me how to think when I see this problem, or show me with another example?

As it would appear that I'm completely stuck on this problem, would it be wise of me to go back to simpler equations and maybe that would refresh some basic knowledge about differential equations that could help me out? I really don't know what to do at this point to understand this.

Edit: I'll look at ehild's suggestion before I go all frustrated on this. And yes, I've obviously seen the formula for the derivative of a fraction. I don't know what I was thinking. I was probably overcomplicating it.

15. Nov 20, 2013

### phyzguy

Well, since you know that:

$$\frac{d}{dx} (f g) =\frac{df}{dx} g + f \frac{dg}{dx}$$

Consider f = y and g = 1/x, and tell us what is:

$$\frac{d}{dx} (y \frac{1}{x})$$

16. Nov 20, 2013

### Staff: Mentor

Solve xy' + y = 3x2, y(1) = 1

Notice that I have a '+' sign on the left side instead of the '-' sign as in your problem.

Looking only at the left side for a moment, I'm trying to work backwards to figure out what <??> should be so that d/dx(<??>) = xy' + y. I am NOT trying to find d/dx(xy' + y)!

Because xy' + y is a sum and the product rule results in a sum, my thinking is that maybe <??> is the product of two factors. Since I see y' maybe one of the things in the product is y.

I notice that d/dx(xy) = xy' + 1y, so xy is what I meant by <??>.

I can now rewrite the differential equation like so:
d/dx(xy) = 3x2

If the derivative of xy is 3x2, then it must be that xy = ∫3x2dx = x3 + C.

Restating this idea in symbols, I have
d/dx(xy) = 3x2
$\Rightarrow$ d(xy) = 3x2dx
$\Rightarrow$ ∫d(xy) = ∫3x2dx
$\Rightarrow$ xy = x3 + C
$\Rightarrow$ y = x2 + C/x

Since y(1) = 1, I have 1 = 12 + C/1, so C = 0
and my solution is y = x2

As mentioned before, it's always a good idea to check your answer. It's easiest to start with the initial condition. Is y(1) = 1? Yes

Does xy' + y = 3x2?
If y = x2, then y' = 2x,
so xy' + y = x(2x) + x2 = 3x2
So, yes, my solution satisfies the differential equation, and I'm done.