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#### LLand314

##### Member

- Mar 26, 2013

- 77

(dy/dx) + y/cosx = 1

e^integral (1/cosx) ?

I feel like this has to do with ln again but not sure

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- Thread starter
- #1

- Mar 26, 2013

- 77

(dy/dx) + y/cosx = 1

e^integral (1/cosx) ?

I feel like this has to do with ln again but not sure

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- Mar 26, 2013

- 77

In your attempt to write the equation in standard linear form, you have made a few errors when dividing through by $\cos(x)$. Try it again and see what you get.

dy/dx + ysinx/cosx = (sinx cosx)/cosx

dy/dx + ytanx = tanx ?

Or would I divide sinx first and move it over the other side before I divide cosx

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Your goal in arranging the equation in standard linear form is to get it in the form:

\(\displaystyle \frac{dy}{dx}+P(x)y=Q(x)\)

and since this equation was given in the form:

\(\displaystyle f(x)\frac{dy}{dx}+g(x)y=h(x)\)

you simply need to divide through by $f(x)$ to get in the the standard form.

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- Mar 26, 2013

- 77

So is my integrating factor 1/cosx ?

Your goal in arranging the equation in standard linear form is to get it in the form:

\(\displaystyle \frac{dy}{dx}+P(x)y=Q(x)\)

and since this equation was given in the form:

\(\displaystyle f(x)\frac{dy}{dx}+g(x)y=h(x)\)

you simply need to divide through by $f(x)$ to get in the the standard form.

So it becomes (dy/dx)(1/cosx) + (ytanx)(1/cosx) = tanx

Integral (ytanx)(1/cosx) = integral tanx

y(1/cosx) = ln|secx| + C

y= cosx ln|secx| + Ccosx ?

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- #6

\(\displaystyle \mu(x)=e^{\int\tan(x)\,dx}=\sec(x)\) and so we have:

\(\displaystyle \sec(x)\frac{dy}{dx}+\sec(x)\tan(x)y=\tan(x)\)

\(\displaystyle \frac{d}{dx}\left(\sec(x)y \right)=\tan(x)\)

\(\displaystyle \int\,d\left(\sec(x)y \right)=\int\tan(x)\,dx\)

\(\displaystyle \sec(x)y=\ln|\sec(x)|+C\)

\(\displaystyle y(x)=\cos(x)\left(\ln|\sec(x)|+C \right)\)