Homework Help: 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.