Question

Calculus 2. Hi I need some help with first-order linear equations and some basic integration. Thanks.

Answer 1

So i have two questions first
i have partially solved (have only solved the dy integral)\[y(x+1)\frac{ dy }{ dx }=x(y^2+1)\]
into \[\int\limits_{?}^{?}\frac{ y }{ y^2+1 }dy=\int\limits_{?}^{?}\frac{ x }{ x+1 }dx\]
i am partially sure that the dx integral would be \[\frac{ x^2 }{ 2 }\] but multiple online integration calulators keep giving me different answers?

Second I would like some help seperating dy and dx here:
\[\frac{ dy }{ dx }= \sqrt{y}\cos ^{2}\sqrt{y}\]
would dx just be 1 because there is no x. thanks.

Answer 2

\[\large\rm \int\limits\frac{x}{x+1}dx\]You're getting x^2/2 from this?
Hmm that's strange :d

Answer 3

I would probably apply a simple algebra trick, adding and subtracting 1 in the numerator,\[\large\rm \int\limits\limits\frac{x}{x+1}dx\quad=\quad \int\limits\frac{x+1-1}{x+1}dx\]And then rewrite it as two separate terms,\[\large\rm =\int\limits\frac{x+1}{x+1}-\frac{1}{x+1}~dx\]First term simplifies,\[\large\rm =\int\limits 1-\frac{1}{x+1}~dx\]

Answer 4

Integrating will give you like an x,
and a log I guess,\[\large\rm =x-\ln|x+1|+c\]Confused by any of that? :o

Answer 5

In regards to your other question,\[\large\rm \frac{dy}{dx}=\left(\sqrt{y}\cos^{2}\sqrt{y}\right)\cdot 1\]Yes, you can think of a 1 being here.
Dividing by the y stuff, moving the dx to the other side gives us,\[\large\rm \frac{dy}{\left(\sqrt{y}\cos^{2}\sqrt{y}\right)}=1\cdot dx\]

Answer 6

Ohh wow you just opened my eyes to a whole new trick (I love easy algebra tricks in calculus :)). In practice would I be able to use this even if it were x+2, could i have subtracted and added 2? Thanks that's a much easier way to do it. Okay and I also see what you mean for the second one. Thanks.

Answer 7

There are a lot of ways to deal with that type of integral.
If you had not been familiar with the trick you would have still had options like
partial fraction decomposition or even just a simple u-substitution (u=x+1).

Yes! Doesn't matter what the other number is. :)

Answer 8

put \[\sqrt{y}=t,y=t^2,\frac{ dy }{ dx }=2t \frac{ dt }{ dx }\]
\[2t \frac{ dt }{ dx }=t \cos ^2 t,2 \frac{ dt }{ dx }=\cos ^2t\]
\[2 \sec ^2t ~dt=dx\]