Question

solve the following equation. I'm wondering if I should be using a table of integral or integration by parts. see attachment

Answer 1

@KingGeorge

Answer 2

@zepdrix

Answer 3

I'm so close...I just got to get the antiderivative on the right

Answer 4

Yeah, I dont think you can do what you did. Im pretty sure you cant have a y in the Q(x) position if youre going to treat this as en equation linear in y.

Answer 5

I was putting it in the form dy/dx + p(x)y = Q(x)
so should I use exact equations like My(x,y)dx+Nx(x,y)dy = 0?

Answer 6

Yeah, I noticed. But you cant have any y in the Q(x) position. But yeah, you might be able to try exact if you can get it set-up in that way.

Answer 7

arghhh got some extra variable in my way that isn't in a dx group or a dy group. it needs to move and join the dy or dx

Answer 8

Try it like a bernoulli.

Answer 9

oh crud I haven't learned that yet...maybe that's why I'm having a hard time

Answer 10

Yeah, it comes out well if done that way. A bernoulli is a variation of a linear equation like you were trying to do, except it comes in the form of
y' + p(x)y = q(x)y^n

In this case, its very easy to set it up like this. In fact, you did have it set up that way before, you just didnt handle it in the proper way.

Answer 11

yeahhh...I should wait for the lecture...I was trying to jump ahead so I can have more study time.

Answer 12

Its not that hard really. The idea is whatever your y expression is in the Q(x) position, you say it's current power is n. So for you, you had 1/y, or basically y^-1 where -1 = n. What you then do is say u = y^(1-n). THis means that your u would be y^2. From here, you multiply everything by du, or basically 2y:

\[2y\frac{dy}{dx} - \frac{4y^{2}}{x}= 2\]What basically happens is you now have the equation linear, but linear in "u" So since u was y^2, I can now rewrite the problem as
\[u' + \frac{ 4u }{ x }=2\]Now you would solve it just like any other linear.

Answer 13

hmmm P(x) = 4/x
e^ integral 4/x
e^ 4 lnx
e^ lnx^4
x^4
Q(x) = 2

Answer 14

Yep yep.

Answer 15

and then I multiply x^4
V(X)dy/dx + P(X)V(X)y = Q(X)V(X)
x^4dy/dx +4/x(x^4)y=2(x^4)