# First order differential, initial value probelm

1. Aug 26, 2010

### illidari

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

In problem 11, determine by inspection at least two solutions of the given initial-value problem.

2. Relevant equations

11. dy/dx = 3y^(2/3) , y(0)=0

3. The attempt at a solution

I can't figure out what they mean by inspection, I have no examples to go off from.

If i make it f(x,y)= 3y^(2/3) then the derivative (df/dy) would be 2/(y^(1/3))
then y could not = 0 which means there is not a guaranteed unique solution.

Now how do I go on to get the solutions :/?

Book says y = 0 and y= x^3 is the answer.

2. Aug 26, 2010

### Pengwuino

Remember, you're looking for a function y(x) that satisfies the differential equation. In other words, some y(x) that, when differentiated with respect to x once, you get 3 * that function to the 2/3 power of the original function.

Surely given y = 0, dy/dx = 3y^(2/3) right? 0^2/3 = 0, d/dx of 0 is 0 right?

3. Aug 26, 2010

### illidari

Am I looking for a function when I take the derivative and plug in 0 for x, I get 0?

Not really understanding how they got the y=x^3

dy/dx = 3x^2

4. Aug 26, 2010

### Staff: Mentor

The differential equation, dy/dx = 3y^(2/3), is separable.
dy/dx = 3y^(2/3) ==>
y^(-2/3)dy = 3dx

Now integrate both sides.

5. Aug 26, 2010

### illidari

That helped thanks.

That method seems to not work on the next problem they gave,

x(dy/dx) = 2y

dy/y = dx/x * 2

ln(y) = 2ln(x)

Book Answer is y = x^2

Is there some other way of solving these ? I am in section 2.1 and separable variables is the name of 2.2. They seem to be explaining how to do #11 in the next section. Any idea what they mean by inspection?

**edit: Forgot about raising e to both sides, that would give me x^2.

Guess this still works :/

6. Aug 26, 2010

### Staff: Mentor

You forgot the constant of integration, which leads me to believe you might also have forgotten it in the previous problem. In the other problem, it turned out to be zero, so your error didn't cost you anything.

7. Aug 26, 2010

### vela

Staff Emeritus
They want you to look at the equation and guess what the answer is based on what you know about differentiation. For example, if you were given the differential equation y'=y, you could guess the answer is ex since you're looking for a function whose derivatives is itself.

For the first problem, y'=3y2/3, you see that differentiating resulted in a factor of 3. When you differentiate x3, the 3 in the exponent comes down, so you might guess that x3 could be a solution. So try it out and see if it works. If it does, you have your solution.

In the second problem, xy'=2y, you could rewrite it slightly as y'=2(y/x). So differentiating introduces a factor of 2 and lowers the power of x by 1. What function do you know is like that?