How do I use substitution to solve a differential equation with a square root?

[User Name]

Homework Statement

Solve the differential equation.

dy/dx = 4x + 4x/square root of (16-x^2)

Homework Equations

Substituting using U...

The Attempt at a Solution

I'm not sure if that's what I am supposed to do, but I tried using the U substitution...

4x + 4x/square root of (16-x^2)
u = 16 - x^2
du = -2x
-2du = 4x (I multiplied both sides by 2 and brought the negative over)

-2 (integration sign [the S]) du + du/square root of u

Take the anti-derivative...

-2 x (u)^(-1/2) + C (changed square root into a power)

-2 x (16-x^2) + C

Unfortunately, that is not the answer they give me. I'm kinda stuck.. am I going in the right direction?

 

[morson]

Everything fell apart right after you made the substitution. Stating that "du = -2x" is incorrect. Rather, du = -2x dx. Then, you find an expression for "dx", sub that in, and integrate with respect to u.

Edit: It's easier to split the integral into a sum of two integrals, so it's the integral of 4x + the other integral.

 

[Schrodinger's Dog]

dy/dx = 4x + 4x/square root of (16-x^2)

I'd imagine that partial fractions would help here.

It would be a lot easier, I'll have a play with it later. Time for work. Just a thought though.

$$\int 4x+=>$$

$$2x^2+C$$

$$\int \frac {4x}{\sqrt(16-x^2)}=>$$

Actually from here just substitute du from here.

 

[HallsofIvy]

No, not partial fractions with that "x" in the numerator.

Let u= 16- x². Then du= -2x dx so that -2du= 4xdx.