How do you solve the differential $dy/dx=4x+(4x)/sqrt(16-x^2)$ ?

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Answer 1 · 2017-09-27T03:49:46

$y = 2x^2 -4sqrt(16-x^2) + C$

Separate variables:

$dy = (4x + (4x)/sqrt(16-x^2)) dx$

Integrate. On the right side, a $u$ substitution can be helpful if the integral cannot be "eyeballed":

$int dy = int (4x + (4x)/sqrt(16-x^2)) dx$
$int dy = int 4x dx + int (4x)/sqrt(16-x^2) dx$
$int dy = int 4x dx + int (4x)(16 - x^2)^(-1/2) dx$

If we let $u = 16 - x^2$ , then $du = -2x dx$ , or conversely:

$2x dx = -du => 4x dx = -2 du$

Substituting and integrating:

$int dy = int 4x dx + int u^(-1/2) (-2du)$
$int dy = int 4x dx -2int u^(-1/2) du$
$y = 2x^2 - 4u^(1/2) + C$

Lastly, substitute back:

$y = 2x^2 - 4(16-x^2)^(1/2) + C = 2x^2 -4sqrt(16-x^2) + C$

How do you solve the differential dy/dx=4x+(4x)/sqrt(16-x^2)dy/dx=4x+(4x)/sqrt(16-x^2)?