0
$\begingroup$

I have the following problem on my assignment: $y''+9y=11\sec^2(3t), 0<t<\frac{\pi}{6}$.

I solve for $r$ and get $\pm 3i$, so then my $y_c=c_1\cos(3t)+c_2\sin(3t)$

Solving for the Wronskian, I get 3.

Therefore, $$u_1=-\int\frac{\sin(3t)11\sec^2(3t)}{3}dt=-\frac{11}{3}\int\tan(3t)\sec(3t)dt=-11\sec(3t)$$ $$u_2=\int\frac{\cos(3t)11\sec^2(3t)}{3}dt=\frac{11}{3}\int\sec(3t)dt=11\ln(|\sec(3t)+\tan(3t)|)$$

Solving for $y_p$ $$y_p=-11\sec(3t)\cos(3t)+11\ln(|\sec(3t)+\tan(3t)|)=-11+11\ln(|\sec(3t)+\tan(3t)|)$$

Then, $y=c_1\cos(3t)+c_2\sin(3t)-11+11\ln(|\sec(3t)+\tan(3t)|)$

However, the program is flagging it as incorrect. I'm not sure where I went astray, and my roommate can't figure out where I went astray either.

$\endgroup$
  • $\begingroup$ @projectilemotion I forgot to add the 11 in my original problem. Too busy trying to figure out mathjax. $\endgroup$ – TobyTobyo Oct 31 '17 at 21:59
  • 1
    $\begingroup$ Your computation of $\int \sec(3t)~dt$ is also wrong. You should have: $$\int \sec(3t)~dt=\color{red}{\frac{1}{3}}\int \sec(u)~du=\frac{1}{3}\ln(|\sec(3t)+\tan(3t)|)+C$$ $\endgroup$ – projectilemotion Oct 31 '17 at 22:03
  • $\begingroup$ My previous comment stated that your computation of $\int \tan(3t)\sec(3t)~dt$ is incorrect. From the above, it seems like you are making consistent mistakes regarding integration by substitution (Possibly the cause of the integration mistakes). Therefore, could you please show us how you would evaluate a simple integral like $\int \cos(3t)~dt$ using the substitution $u=3t$? $\endgroup$ – projectilemotion Oct 31 '17 at 22:45
2
$\begingroup$

Check your constants

$$ -{11\over 3}\int \sec 3t \tan 3t \, dt = -{11\over 9} \sec 3t $$

$$ {11\over 3}\int \sec 3t \, dt = {11\over 9} \ln |\sec 3t + \tan 3t| $$

Remember that when you have something like $$\int f(at)dt$$

The anti-derivative should be divided by the constant, not multiplied $${1\over a}F(at) + C$$

This effect is sort of like a reverse chain rule $$ F'(at) = af(at)$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.