Problem 39 - Essential Calculus

Question

Find the particular solution of the differential equation. $$ x \frac{d y}{d x}=\sqrt{x^{2}-9}, \quad x \geq 3, \quad y(3)=1 $$

Step-by-Step Solution

Answer

The particular solution of the given differential equation is \(y = \frac{\sinh 2 \cosh^{-1}(\frac{x}{3})}{4} - \frac{\cosh^{-1}(\frac{x}{3})}{2} + 1\).

Step 1: Rewrite the equation

Rewrite the equation in the separable form. \[x \frac{d y}{d x} = \sqrt{x^{2}-9} \implies \frac{d y}{dx} = \frac{\sqrt{x^{2}-9}}{x}\]This equation is separable and can be written as \(dy = \frac{\sqrt{x^{2}-9}}{x} dx\).

Step 2: Integrate Both Sides

Integrate both sides to find the function \(y\). \[\int dy = \int \frac{\sqrt{x^{2}-9}}{x} dx\]The left side is quite straightforward and will simply result in \(y\). The right side requires knowledge of integral calculus and can be solved by substitution. Let \(x = 3\cosh t\), then \(dx = 3\sinh t dt\) and \( x\geq3 \rightarrow t \geq 0\). Then the integral becomes \[\int dy = \int \frac{\sqrt{(3\cosh t)^{2}-9}}{3\cosh t} \times 3\sinh t dt = \int \sqrt{\cosh^{2}t-1} \sinh t dt = \int \sinh^{2} t dt\]The integral of \(\sinh^{2} t\) can be simplified as\[\int \sinh^{2} t dt = \int \frac{\cosh 2t -1}{2} dt = \frac{\sinh 2t}{4} - \frac{t}{2} + C\]After substituting \(t\) back in (remembering that \( t = \cosh^{-1}(\frac{x}{3}) \)), the integral becomes \[\int dy = \frac{\sinh 2 \cosh^{-1}(\frac{x}{3})}{4} - \frac{\cosh^{-1}(\frac{x}{3})}{2}+ C\]

Step 3: Solve for the Constant

The next step is to solve for the constant \(C\). Given that \(y(3) = 1\), we can substitute these values into the equation to get \(C\). \[1 = \frac{\sinh 2 \cosh^{-1}(\frac{3}{3})}{4} - \frac{\cosh^{-1}(\frac{3}{3})}{2} + C\]Which simplifies to \(1 = \frac{0}{4} - 0 + C\), then we find that \(C = 1\).

Step 4: Write the Particular Solution

Substituting \(C = 1\) into our integral from Step 2, we can write the particular solution for the differential equation.\[y = \frac{\sinh 2 \cosh^{-1}(\frac{x}{3})}{4} - \frac{\cosh^{-1}(\frac{x}{3})}{2} + 1\]