# Little help with differential equations

1. Feb 18, 2009

### Titans86

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

$$\frac{dy}{dx} = \sqrt{xy^3} , y(0) = 4$$

3. The attempt at a solution

So;

$$\frac{dy}{dx} = x^{\frac{1}{2}}y^{\frac{3}{2}} \Rightarrow \int y^{-\frac{3}{2}}dy = \int x^{\frac{1}{2}}dx \Rightarrow -2y^{-\frac{1}{2}} = \frac{2}{3}x^{\frac{3}{2}} + C \Rightarrow y^{-\frac{1}{2}} = -\left(\frac{1}{3}x^{\frac{3}{2}} + \frac{1}{2}C\right) \Rightarrow y^{\frac{1}{2}} = -\left(\frac{1}{\frac{1}{3}x^{\frac{3}{2}} + \frac{1}{2}C}\right) \Rightarrow y = \frac{1}{(\frac{1}{3}x^{\frac{3}{2}} + \frac{1}{2}C)^2}$$

Then Putting in the conditions mentioned above:

$$4 = \frac{1}{(0 + \frac{1}{2}C)^2} \Rightarrow \frac{1}{4}C^2 = \frac{1}{4} \Rightarrow C = 1$$

Yet my book shows $$C = \frac{3}{2}$$

2. Feb 18, 2009

### Staff: Mentor

Your work looks fine to me except near the end. C can also equal -1.
Is the book's answer y = 1/[1/3*x^(3/2) + 1/2 * 3/4]^2? (I.e., same as yours but with C = 3/2 rather than C = 1 as you have it.)

I don't see why your general solution wouldn't be a solution to the DE, and I don't see any problems with your use of the initial condition to find the particular solution. I would want to verify that the book's solution solves the DE and initial condition. After all, math text occasionally have typos.

3. Feb 18, 2009

### Titans86

you mean there can be more then one particular solution?