Question

in exercises 1 - 10, find the general solution to the exact differential equation.

1. $\frac{dy}{dx}=5x^{4}-\sec^{2}x$

Explanation:

Step1: Integrate both sides with respect to \(x\)

$$y=\int(5x^{4}-\sec^{2}x)dx$$

Step2: Integrate term - by - term

For \(\int5x^{4}dx\), using the power rule \(\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n
eq - 1)\), we have \(\int5x^{4}dx=5\times\frac{x^{4+1}}{4 + 1}=x^{5}\).
For \(\int\sec^{2}x dx\), since \(\frac{d}{dx}(\tan x)=\sec^{2}x\), then \(\int\sec^{2}x dx=\tan x\).
So \(y=\int(5x^{4}-\sec^{2}x)dx=\int5x^{4}dx-\int\sec^{2}x dx=x^{5}-\tan x + C\)

Answer:

\(y=x^{5}-\tan x + C\)