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

Question

Luke Alvarez · Accepted Answer

#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#

Emma Aldridge · Answer

undefined To solve the differential equation $\frac{dy}{dx} = 4x + \frac{4x}{\sqrt{16 - x^2}}$, you can rewrite it as:

$\frac{dy}{dx} = 4x + \frac{4x}{\sqrt{16 - x^2}} = 4x + \frac{4x}{\sqrt{(4)^2 - x^2}}$

This suggests that we can use the substitution $x = 4\sin(\theta)$ to simplify the expression. Then, $dx = 4\cos(\theta)d\theta$. Substituting these into the differential equation, we get:

$\frac{dy}{d\theta} = 4(4\sin(\theta)) + \frac{4(4\sin(\theta))}{\sqrt{(4)^2 - (4\sin(\theta))^2}}$

Simplify this and integrate to find $y(\theta)$, and then substitute back $x = 4\sin(\theta)$ to find $y(x)$.