Integration by substitution diff. eq.

1. Oct 29, 2012

lordofpi

1. The problem statement, all variables and given/known data
From Larson, 9th Edition: Section 4.5. Solve the differential equation

$$\frac{\operatorname{d}y}{\operatorname{d}x}=4x+ \frac{4x}{\sqrt{16-x^2}}$$

2. Relevant equations

3. The attempt at a solution
Well, I can get my book's answer, but not through doing things in the prescribed way, because I get a different answer following the section's guidelines and examples. Using those, I choose a variable $u=16-x^2$. Taking the derivative, we obtain $du=2x \ \operatorname{d}x$. Rearranging this to get a $4x$ factor, we have $$-2 \operatorname{d}u = 4x \operatorname{d}x$$.

Plugging this back into the differential equation and rearranging the d$x$ gives us:
$$\int \operatorname{d}y = -2 \int \operatorname{d}u + (-2)\int u^\frac{-1}{2} \operatorname{d}u$$

Now I wind up with $$y=-2u - 4\sqrt{u} + C$$
Which, using substitution becomes $$2x^2-4\sqrt{16-x^2}-32+C$$, but the book's answer is $$2x^2 -4\sqrt{16-x^2} + C$$

Now I can get this answer if I just take the integral of the first $4x$ and only use substitution for the second term, but I am not sure why my answers should be different. Unless it really doesn't matter and the 32 from my answer just becomes part of the constant of integration $C$

Last edited: Oct 29, 2012
2. Oct 29, 2012

Dick

It's exactly what you think. The -32 can be absorbed into the C. Indefinite integrals can always have a constant added and they work just as well.

3. Oct 29, 2012

lordofpi

Thanks Dick. Just out of curiosity, what would happen if this were a definite integral? The same, I am assuming, since we would have to subtract -32 from -32?

4. Oct 29, 2012

Dick

You wouldn't see the -32 in a definite integral, because again, as you say, it would cancel. That's why it's not important.

5. Oct 30, 2012

SteamKing

Staff Emeritus
It's not necessary to use u-substitution to evaluate both integrals simultaneously.
The integral of 4x is readily evaluated using the standard formula. The u-substitution
can be used to evaluate the integral with the radical.