# Differential Equation expression help

1. Aug 3, 2010

### Moonflower

Here's the question:

Let f be the function satisfying f'(x)=x$$\sqrt{f(x)}$$ for all real numbers where f(3)=25.

a. Find f''(3)

b. Write an expression for y-f(x0 by solving the differential equation $$\frac{dy}{dx}$$ = x$$\sqrt{y}$$ with the initial condition of f(3)=25.

For a, I got $$\frac{x^2}{2}$$+$$\sqrt{f(x)}$$, so my answer was $$\frac{19}{2}$$.

For b, I immediately substituted, getting dy/dx=3sqrt(25). then, dy/dx=15 -> dy=15dx -> integrate, y=15x+c, and since the initial condition is f(3)=25, by substitution, C=-20. My answer in the end was y=15x-20.

Am I on the right track? Thanks.

2. Aug 3, 2010

### Mentallic

I haven't actually studied differential equations, so I don't know any methods to solve them, and this is coming from my own intuition so I advise you to read with caution

$$\frac{dy}{dx}=x\sqrt{y}$$

$$y=\left(\frac{x^2}{4}+c}\right)^2$$

would satisfy this differential.

I don't quite understand what part b) is asking though. Part a) is pretty easy if you know what the function is.

3. Aug 3, 2010

### Moonflower

How did you get to that result? It satisfies, but I couldn't reach that result..

4. Aug 3, 2010

### Gib Z

Separation of Variables. Also, your expression for the second derivative was correct, but 19/2 isn't so check your arithmetic.

5. Aug 3, 2010

### Moonflower

well, 9/2 + 5 is 19/2, or 9.5 isn't it?

Also, can you describe me the process?

thanks.

6. Aug 3, 2010

### Gib Z

Ahh foolish me, I put in x=5 rather than x=3, you are correct.

To separate variables, start with $$\frac{dy}{dx}=x\sqrt{y}$$ and then take over the $\sqrt{y}$ term to give $$\frac{1}{\sqrt{y}}\frac{dy}{dx} = x$$, then integrate both sides with respect to x.

7. Aug 3, 2010

### Mentallic

This part is important to note:
I just realized that to get the square root of the function and have an x multiplied by it, I need a quadratic of the form ax2+c taking all to the 2nd power since we are going to minus one from the power, which is also the square root.

So we have $$y=(ax^2+c)^2$$ and then $$y'=2(ax^2+c).2ax=4ax(ax^2+c)$$

Now we need a=1/4 to satisfy the problem.

8. Aug 3, 2010

### Mentallic

Oh and Gib Z's method works much better for those that don't simply notice it! It did take longer to solve, but there, I've learnt differentials.