Integration by substitution diff. eq.

[lordofpi]

Homework Statement

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}}

Homework Equations

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 dx gives us:
<br /> \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

 

[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.

 

[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?

 

[Dick]

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?

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.

 

[SteamKing]

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.