Question

Sove the IVP

xy' + y = 0 y(4) = 6

Answer 1

@zepdrix
I did this:
\[y'=-y/x\\y'=-u\\ y=-u^2/2 + c\]
Did I make a mistake with the integration part?

Answer 2

Oh did they want you to make a substitution again?

Answer 3

Yah looks ok so far.
If the instructions didn't ask you to make a substitution, I would avoid doing so.

Because you're going to end up with an answer of \(\large y=f(x,y)\)
x's and y's on the right.

And this is a nice easy problem that can be solved in terms of \(\large y=f(x)\)

Answer 4

or just use reverse product
\[xy' + y = 0\]
\[\frac{d}{dx}(xy)=0\]
\[xy=\int\limits 0dx\]

Answer 5

\[\large y'=-\frac{y}{x}\qquad\rightarrow\qquad \frac{1}{y}dy=-\frac{1}{x}dx\]

Answer 6

Ah yes, that works nicely too c: what sam did.

Answer 7

Oh I see, I should rather just do the normal separable method thing...

Answer 8

so then I would get\[\ln|y| = -\ln|x| + c_1\\y=-x+c_2\]

And with the initial values
c_2 = 10
and
y=-1+10
?

Answer 9

ah
y=-x+10

Answer 10

Woops after you integrated, let's be careful with the next step.

Answer 11

\(\large \ln y=-\ln x+c \qquad \rightarrow \qquad \ln y=\ln\left(\dfrac{1}{x}\right)+c\)

Exponentiating both sides gives us,\[\large y=e^{\left(\ln\left(\dfrac{1}{x}\right)+c\right)} \qquad \rightarrow \qquad y=C\frac{1}{x}\]

Answer 12

That unknown constant is now a factor, it's no longer being added, due to the exponentiation.

I can show another step in there if you're confused by the last part.

Answer 13

Thanks, I'm not confused yet haha.
I will be on the lookout for this kind of ln thing, I tend to do what I just did...

Answer 14

Yah it's a very easy mistake to make :)
Make sure you exponentiate EVERYTHING on the right! heh

Answer 15

Thanks!

In this problem, would we get y=24/x?

Answer 16

Ah yes, that sounds right.

Answer 17

@zepdrix
I did another problem which led me to this:
\[\ln|y|=2x^2-\ln|\cos x| +C\]
After this is exponentiated, how do I simplify the equation?

Answer 18

\[\large y=e^{\left(2x^2-\ln|\cos x| +c\right)}\]
\[\large y=e^{(2x^2-\ln(\cos x))}\cdot e^c\]
\[\large y=Ce^{(2x^2-\ln(\cos x))}\]

Hmm like that I suppose.

Answer 19

I think I know... I should try to combine the ln's?
So,\[\ln|y| + \ln|\cos x|=2x^2 +C\\ \ln|y \cos x|=2x^2 + C \\ y \cos x = c_2 e^{2x^2} \\ y = c_2\frac{e^{2x^2}}{\cos{x}}\]?

Answer 20

Oh Yes i missed a step in there, my bad.
\[\large y=Ce^{(2x^2+\ln(\sec x))}\]\[\large y=Ce^{2x^2}\cdot e^{\ln \sec x}\]\[\large y=C (\sec x) e^{2x^2}\]

Your method seems to have worked also :) Nice

Answer 21

Thanks zep!