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#### find_the_fun

##### Active member

- Feb 1, 2012

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So \(\displaystyle \frac{dy}{dx}+\frac{sinx}{cosx}y=\frac{1}{cosx}\)

\(\displaystyle \frac{dy}{dx}+tan(x)y=csc(x)\) therefore \(\displaystyle P(x)=tan(x)\)

Let \(\displaystyle \mu (x) = e^{\int P(x) dx}=e^{\int tan(x) dx} =\frac{1}{cosx}\)

multiply both sides of the equation by integrating factor

\(\displaystyle \frac{dy}{dx} \frac{1}{cosx}+y\frac{tanx}{cosx}=\frac{1}{cos^2x}\)

\(\displaystyle \frac{d}{dx}[\mu(x)y]=\frac{d}{dx}[\frac{1}{cosx}y]\) use product rule \(\displaystyle (\frac{1}{cosx})'y+\frac{1}{cosx}y'=\frac{tanx}{cosx}y+\frac{1}{cosx}\frac{dy}{dx}\)

**Question 1**: how come when we use the product rule and take the derivative of y with respect x we get \(\displaystyle \frac{dy}{dx}\) and not 0? I say 0 because the derivative of a constant is 0 and aren't we treating y as a constant?

Next step: integrate

\(\displaystyle \int \frac{d}{dx}[\frac{1}{cosx}y]dx = \int \frac{1}{cosx} x\)

\(\displaystyle \frac{y}{cosx}=\ln{|-cosx|}+C\)

and \(\displaystyle y=\ln{|-cosx|}cosx+Ccosx\)

**Question 2**: the back of book has \(\displaystyle y=sinx+Ccosx\) Where did sin(x) come from?