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lordofpi
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Homework Statement
From Larson, 9th Edition: Section 4.5. Solve the differential equation
[tex]\frac{\operatorname{d}y}{\operatorname{d}x}=4x+ \frac{4x}{\sqrt{16-x^2}}[/tex]
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 [itex]u=16-x^2[/itex]. Taking the derivative, we obtain [itex]du=2x \ \operatorname{d}x[/itex]. Rearranging this to get a [itex]4x[/itex] factor, we have [tex]-2 \operatorname{d}u = 4x \operatorname{d}x[/tex].
Plugging this back into the differential equation and rearranging the d[itex]x[/itex] gives us:
[tex]
\int \operatorname{d}y = -2 \int \operatorname{d}u + (-2)\int u^\frac{-1}{2} \operatorname{d}u[/tex]
Now I wind up with [tex]y=-2u - 4\sqrt{u} + C[/tex]
Which, using substitution becomes [tex]2x^2-4\sqrt{16-x^2}-32+C[/tex], but the book's answer is [tex]2x^2 -4\sqrt{16-x^2} + C[/tex]
Now I can get this answer if I just take the integral of the first [itex]4x[/itex] 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 [itex]C[/itex]
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