Question

Use the differential equation and the specified initial condition to find \(y .\) $$ \begin{array}{l} \frac{d y}{d x}=\frac{1}{4+x^{2}} \\ y(2)=\pi \end{array} $$

Short answer

The solution to the differential equation is \( y(x) = \frac{1}{2}\tan^{-1}(\frac{x}{2}) + \frac{7π}{8} \)

Step-by-step solution

Integrate the given differential equation

Integrate \(\frac{d y}{d x}=\frac{1}{4+x^{2}}\) with respect to \(x\) on both sides, which implies that the equation will become: \(y(x) = \int \frac{1}{4+x^{2}} dx + C\). Here, \(C\) is the constant of integration.

Evaluate the integral

The integral \(\int \frac{1}{4+x^{2}} dx\) can be evaluated using the formula \(\int \frac{1}{a^{2}+x^{2}} = \frac{1}{a} \tan^{-1}(\frac{x}{a}) + C\), where \(a=2\) (since \(a^{2}=4\)). So, \(y(x) = \frac{1}{2} \tan^{-1}(\frac{x}{2}) + C\).

Use the given initial condition y(2)=π to determine C

Substitute \(x=2\) and \(y(2)=\pi\) in the equation from Step 2: \(\pi = \frac{1}{2} \tan^{-1}(1) + C\). Now, \(\tan^{-1}(1) = \frac{π}{4}\), therefore, \(C = \pi - \frac{π}{8} = \frac{7π}{8}\).

Write down the final solution

Replace constant \(C\) in the equation from Step 2 with the value we got, so the solution of the differential equation is: \(y(x) = \frac{1}{2}\tan^{-1}(\frac{x}{2}) + \frac{7π}{8}\)