Question

Find the particular solution of the differential equation. $$ \sqrt{x^{2}+4} \frac{d y}{d x}=1, \quad x \geq-2, \quad y(0)=4 $$

Short answer

The particular solution to the given differential equation is \(y(x) = \sinh^{-1}(x/2) + 4\), valid for \(x \geq -2\).

Step-by-step solution

Conversion to Standard Form

Convert the given differential equation \(\sqrt{x^{2}+4} \frac{d y}{d x}=1\) into the standard form used to define ordinary differential equations. You get \(\frac{d y}{d x}=\frac{1}{\sqrt{x^{2}+4}}\).

Integration of the Differential Equation

Calculate the integral of both sides of the equation. This results in \(y(x) = \int \frac{1}{\sqrt{x^{2}+4}} dx\). Calculating this integral gives \(y(x)= \sinh^{-1}(x/2) + C\), where \(C\) represents the constant of integration.

Apply Initial Condition and Solve for C

Substitute the initial conditions \(x=0, y=4\) into the equation \(4 = \sinh^{-1}(0/2) + C\) and solve for \(C\). You get \(C = 4\).

Formulation of the Particular Solution

Substitute \(C\) into the equation \(y(x)\) to get your particular solution. You get the final answer as \(y(x) = \sinh^{-1}(x/2) + 4\) when \(x \geq -2\).