- #1
Vuldoraq
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Homework Statement
x[tex](\theta)[/tex]= x tanθ−cosθ
Homework Equations
The integrating factor is (note it should be negative tan, but the latex won't display it);
[tex]I= \int -tan \theta\ d \theta[/tex]
The solution is given by;
[tex]x \ e^{I}=-\int \ e^{I}*cos\theta\ d \theta[/tex]
The Attempt at a Solution
[tex]I=\int \tan \theta\ d \theta[/tex]
[tex]=-(-\ln|cos \theta |)[/tex]
So, [tex]e^{I}=cos \theta[/tex]
Therefore,
[tex]x*cos \theta = \int \ cos^{2} \theta d \theta[/tex]
Using the half angle formulae gives,
[tex]x*cos \theta =\frac{1}{2} \int \cos(2 \theta) +1[/tex]
Which, after integration and simplification gives,
[tex]x=-\frac{1}{2}*(sin \theta + \theta\ sec\theta)[/tex]
When I check my answer I end up with zero,
[tex]\frac{dx}{d \theta}=-\frac{1}{2}*(cos \theta \ +\ sec\theta \ +\ \theta*sec \theta \tan \theta)[/tex]
[tex]x*tan \theta=-\frac{1}{2}*(sin \theta * tan \theta + \theta* sec \theta* tan \theta)[/tex]
[tex]\frac{dx}{d \theta}-x*tan \theta=\frac{1}{2}*[(\frac{cos^{2}\theta\ {-1}}{cos \theta}-\theta* sec \theta* tan \theta)-(-\frac{sin^{2} \theta}{cos \theta}-\theta* sec \theta* tan \theta)][/tex]
[tex]=\frac{sin^{2} \theta \ - \ (1-cos^{2} \theta)}{2* cos \theta}=0[/tex]
Please could someone go through my calculation and see where my mistake is?
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