Integration of a differential equation

Dec 2008
105
3
Hi,

I am unsure if I am getting the correct answer to my problem, I have to find y

\(\displaystyle \frac{dy}{dx} = y\cos(x)\) which goes to

\(\displaystyle \int\frac{1}{\cos(x)\cdot dx} = \int\frac{y}{dy}\) then

\(\displaystyle ln(\cos(x)) + c = y^2\) therefore

\(\displaystyle y = \sqrt{ln(\cos(x)) + c}\)

Is this correct, if not where am i going wrong?

Thanks
 
Dec 2009
872
381
1111
Hi,

I am unsure if I am getting the correct answer to my problem, I have to find y

\(\displaystyle \frac{dy}{dx} = y\cos(x)\) which goes to

\(\displaystyle \int\frac{1}{\cos(x)\cdot dx} = \int\frac{y}{dy}\) then

\(\displaystyle ln(\cos(x)) + c = y^2\) therefore

\(\displaystyle y = \sqrt{ln(\cos(x)) + c}\)

Is this correct, if not where am i going wrong?

Thanks
Dear Beard,

\(\displaystyle \frac{dy}{dx} = y\cos~x\)

\(\displaystyle \frac{1}{y}\frac{dy}{dx}=cos~x\)

Now integrate both sides with respect to x,

\(\displaystyle \int{\frac{1}{y}\frac{dy}{dx}}dx=\int{cos~x}~dx\)

\(\displaystyle \int{\frac{1}{y}}dy=sin~x+C\) ; Cis an arbitary constant.

\(\displaystyle ln\mid{y}\mid=sin~x+C\)

Hope this will help you.
 
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Dec 2008
105
3
Nice and clear to follow thankyou
 

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
And never, ever again write "\(\displaystyle \int \frac{f(x)}{dx}\)"!!!
 
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