Question

solve the given differential equation. (y^{2}+4)dx = ysec^{2}(x)dy

Explanation:

Step1: Separate variables

Separate the variables in the differential equation \((y^{2}+4)dx = y\sec^{2}(x)dy\) to get \(\frac{dx}{\sec^{2}(x)}=\frac{y}{y^{2}+4}dy\). Since \(\frac{1}{\sec^{2}(x)}=\cos^{2}(x)\), the equation becomes \(\cos^{2}(x)dx=\frac{y}{y^{2}+4}dy\).

Step2: Integrate both sides

Recall that \(\cos^{2}(x)=\frac{1 + \cos(2x)}{2}\). Integrate the left - hand side \(\int\cos^{2}(x)dx=\int\frac{1+\cos(2x)}{2}dx=\frac{1}{2}\int(1 + \cos(2x))dx=\frac{1}{2}(x+\frac{\sin(2x)}{2})+C_1=\frac{x}{2}+\frac{\sin(2x)}{4}+C_1\). For the right - hand side, let \(u = y^{2}+4\), then \(du = 2y\;dy\) and \(\int\frac{y}{y^{2}+4}dy=\frac{1}{2}\int\frac{du}{u}=\frac{1}{2}\ln|u|+C_2=\frac{1}{2}\ln|y^{2}+4|+C_2\).

Step3: Combine the results

We have \(\frac{x}{2}+\frac{\sin(2x)}{4}=\frac{1}{2}\ln|y^{2}+4|+C\) (where \(C = C_2 - C_1\)).

Answer:

\(\frac{x}{2}+\frac{\sin(2x)}{4}=\frac{1}{2}\ln|y^{2}+4|+C\)