# Thread: FACULTY OF MECH ENGINEERING exam #2

1. ## FACULTY OF MECH ENGINEERING exam #2

Hello, im new to the forum and this is my debut thread/problem. I am currently a student enrolling in mechanical engineering at the University of Ljubljana, Slovenia. Anyways we had an exam a few days ago and it was suprising to me that only 2 people managed to get above 50% out of 100 students. The differential equation that gave me problems was this:

f(x)=ln(tg(2x))

Prove that: f "(x)+4ctg(4x)*f '(x)=0

*=product
ln=natural logarithm
tg=tangent
ctg=cotangent

2. $f(x)=\ln(\tan(x))$

Using the chain rule find $f'(x)$ and $f''(x)$

and check them in the equation you have.

3. yeah i did that but its very complicated.
i got f '(x)= 2/(cos^2 (2x) * tg(2x))
and f ''(x)= -4 sin(2x)*tg(2x)-2*(cos^2 (2x) - 1/cos^2 (2x))/ (cos^4 (2x)*tg^2 (2x))

4. Originally Posted by spanky489

f(x)=ln(tg(2x))

Prove that: f "(x)+4ctg(4x)*f '(x)=0

*=product
ln=natural logarithm
tg=tangent
ctg=cotangent
$f'(x) = \frac{2\sec^2(2x)}{\tan(2x)} = \frac{2}{\cos^2(2x) \tan(2x)} =$ $\frac{2}{\sin(2x)\cos(2x)} = \frac{4}{2\sin(2x)\cos(2x)} = 4\csc(4x)$

$\textcolor{red}{f'(x) = 4\csc(4x)}$

$\textcolor{blue}{f''(x) = -16\csc(4x)\cot(4x)}$

$\textcolor{blue}{-16\csc(4x)\cot(4x)} + 4\cot(4x) \cdot \textcolor{red}{4\csc(4x)} = 0$

5. $y''+4\cot(4x)y'=0$

$\frac{dy'}{y'}=-4\cot(4x)dx$

$\log(y')=-\log(\sin(4x))+C$

$y'=\frac{k}{\sin(4x)}$

$y=\frac{k}{4}\log(tan(2x))+c_2$

so $y(x)=\log(tan(2x)$ is a particular solution.

6. Originally Posted by shawsend
$y''+4\cot(4x)y'=0$

$\frac{dy'}{y'}=-4\cot(4x)dx$

$\log(y')=-\log(\sin(4x))+C$

$y'=\frac{k}{\sin(4x)}$

$y=\frac{k}{4}\log(tan(2x))+c_2$

so $y(x)=\log(tan(2x)$ is a particular solution.
hehe thanks! but i think this is too complex for 1st semester math. anyways thx for help guys im off to bed gotta wake up early tomorrow and protest my exam score. thx again!