Problem 23 - Essential Calculus

Question

Solve the differential equation. $$ \frac{d y}{d x}=4 x+\frac{4 x}{\sqrt{16-x^{2}}} $$

Step-by-Step Solution

Answer

The solution to the given differential equation is \( y = 2x^2 - 4\sqrt{16-x^2} + C \)

Step 1: Identify the type of differential equation

The given equation is \(\frac{d y}{d x}=4 x+\frac{4 x}{\sqrt{16-x^{2}}}\) which is a first-order non-homogeneous ordinary differential equation. We are tasked with finding a solution for \( y \) in terms of \( x \).

Step 2: Integrate both sides of the equation

We obtain the original function \(y\) by integrating both sides of the equation ´dy/dx = 4x + 4x/(sqrt(16 - x²))´ with respect to \(x\). This gives us \( y = \int (4x + \frac{4x}{\sqrt{16-x^2}}) dx \)

Step 3: Compute the integral

Separate the integral into two parts to simplify the computation. \( y = \int 4x dx + \int \frac{4x}{\sqrt{16-x^2}} dx \). The first integral simplifies to \(2x^2\). The second integral is a standard form whose solution is \(-4\sqrt{16-x^2}\). Thus, the solution to the initial differential equation is \( y = 2x^2 - 4\sqrt{16-x^2} + C \), where \( C \) is the integration constant.