# Differential Equation (3)

• der.physika

#### der.physika

Solve the Differential Equation $$2yy\prime\prime+2xy\prime=0$$

set $$p=y\prime$$, and then it becomes case 1 in the textbook

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Doesn't that simplify to y=-x?

Doesn't that simplify to y=-x?

Did you consider the
$$y\prime$$ that it's not the same as $$y$$

Did you consider the y' that it's not the same as y
What's your point? y' is a common factor in the equation you wrote. Did you not post the problem correctly?
Solve the Differential Equation $$2yy\prime+2xy\prime=0$$

set $$p=y\prime$$, and then it becomes case 1 in the textbook
You do realize we don't know what textbook you're using and even less likely to have a copy, right?

What's your point? y' is a common factor in the equation you wrote. Did you not post the problem correctly?

You do realize we don't know what textbook you're using and even less likely to have a copy, right?

Okay, objection 1 makes no sense... but I'll give you you're right for objection 2.

Here's what the textbook says for case 1:

$$y\prime=p, y\prime\prime=p\prime\mbox {Dependent variable y missing}$$

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Okay, objection 1 makes no sense
Really? You don't see how x=-y is a solution to 2yy'+2xy'=2y'(x+y)=0?

Really? You don't see how x=-y is a solution to 2yy'+2xy'=2y'(x+y)=0?

How did you get that solution besides canceling things out?

Cause, this is like a very annoying class, like what method did you use? Did you use $$p\equiv{y}\prime$$

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You get that solution by canceling things out. It's not the only solution to the original equation, however.

You get that solution by canceling things out. It's not the only solution to the original equation, however.

could you possibly do it, by using the substitution I posted?

I don't see what that substitution buys you. Also, it's not your textbook's case 1 as y appears in the differential equation.

ok I posted this wrong, i feel stupid

$$2yy\prime\prime+2xy\prime=0$$

how do I solve this?

give me a hint maybe?

I don't know. Maybe someone else here has a suggestion.

What's the point of the 2's? I would have just canceled them out from the beginning.

(Y*Y'')/Y' = -X

(Y''/Y')(Y) = -X

Set P = Y', then:

(P'/P)Integral(P) = -X

Ok, I'm sorry. This is going nowhere. Given me 24 hours to think about it.