# Simple diff eq that i cant figure out

1. Aug 30, 2007

### boomer22

1. The problem statement, all variables and given/known data
y ' [x] = xy
y[0] = 1

2. Relevant equations
none

3. The attempt at a solution

I posted this somewhere else and 2 people got the answer is y = e^x

I am getting something else though.

I get up to here:

dy/dx = xy

1/y dy = x dx

ln y = (x^2)/2 + C

that is what i get to, but here is what the other person got from this point:

y = e^[ (x^2/2) + C ] (Im not confused on e^ln^y = y fyi)

then he simplified that to:

y = Ce^x (I DONT understand that property of natural logs)

1 = C * e^0

C = 0

y = e^x

anyone care to explain? dont worry if you know a better way and start from scratch, ignore mine and this other guys answer if you want.

2. Aug 30, 2007

### VietDao29

Yup, so far, so good.

Yup, this is correct. Can you get this step? We have:

$$\ln y = x \Rightarrow e ^ {\ln y} = e ^ x \Rightarrow y = e ^ x$$

Nope, this is wrong. You can simply check it by differentiate y with respect to x. If y = ex, then, y' = ex = y, not xy.

Here's some properties of exponential that you should remember:
(1). $$\alpha ^ {\beta + \gamma} = \alpha ^ \beta \times \alpha ^ \gamma$$

(2).$$\alpha ^ {\beta - \gamma} = \frac {\alpha ^ \beta}{\alpha ^ \gamma}$$

(3).$$\alpha ^ {-\gamma} = \alpha ^ {0 - \gamma} = \frac{\alpha ^ 0}{\alpha ^ \gamma} = \frac{1}{\alpha ^ \gamma}$$

(4).$$\alpha ^ {\beta \times \gamma} = \left( \alpha ^ \beta \right) ^ \gamma$$

(5).$$\alpha ^ {\frac{\beta}{\gamma}} = \sqrt[\gamma]{\alpha ^ \beta}$$

$$y = e ^ {\frac{x ^ 2}{2} + C}$$

Using (1), we have:

$$\Rightarrow y = e ^ C \times e ^ {\frac{x ^ 2}{2}}$$ since C is a constant, we have eC is also a constant, we denote it D, we have:

$$\Rightarrow y = D \times e ^ {\frac{x ^ 2}{2}}$$

Can you go from here? Can you find D? :)

Last edited by a moderator: Aug 30, 2007
3. Aug 30, 2007

### boomer22

haha no, i cant actually. please continue.

4. Aug 30, 2007

### nicktacik

Use the condition y(0) = 1 to find D.

5. Aug 30, 2007

### boomer22

ok, D = 0 then

6. Aug 30, 2007

### d_leet

No, because that forces y=0 for all x, which certainly does not satisfy y(0)=1.

7. Aug 31, 2007

### HallsofIvy

In $$y= De^{\frac{x^2}{2}}$$, replace y with 1, replace x with 0, and solve for D.

8. Sep 3, 2007

### l46kok

Just so you don't get thrown off

y = e^[ (x^2/2) + C ]

does not simplify down to

y = Ce^x (What happened to the x^2 then?)

9. Sep 3, 2007

### HallsofIvy

Yes, that was pointed out before.
$$y= e^{\frac{x^2}{2}+ C$$
does simplify to
$$y= Ce^{\frac{x^2}{2}$$
But this new "C" is not the same as the previous "C"!

Now if y= 1 when x= 0, that becomes
[tex]y= 1= Ce^{0}[/itex]
Solve that for C.

10. Sep 6, 2007

### boomer22

ok, C = 1 then

what is next?

11. Sep 6, 2007

### bob1182006

plug in C=1 into y=Ce^[(x^2)/2]