2nd Order Differential Equation help

Not sure how to approach this

Use the substitution they've suggested to rewrite the original equation in terms of dv/dx, v and x. You then have a 1st order equation which you know how to solve

Yeah but if I were to get dv/dx I would have to use the product rule giving me

dv/dx = dv/dy x dy/dx

and i don't know what dv/dx or dv/dx is equal to

leaving me with

dv/dx = vdv/dy

I don't understand what you're doing there - you don't need any derivatives w.r.t.y!

You are told to set $v = \dfrac{dy}{dx}$, so $\dfrac{dv}{dx} = \dfrac{d^2y}{dx^2}$. Now just put these into your DE which becomes a simpler DE involving v and x.

But then you'd be left with

xdv/dx -2v + x = 0

And you can seperate the variables with that

I agree

This is a standard 1st order DE which you can use the integrating method to solve.

Oops sorry, I meant can't :P

But then you'd be left with

xe^-x^2dv/dex - 2ve^-x^2 = -xe^-x^2

Which can't be tranformed in d/dx(...) = -xe^-x^2 because of the v and x (you'd be left with e^-x^2 and something else

I think you need to check your integrating factor - I don't get any exponentials in mine

I thought the integrating factor is usually e^integratePx

I think you need to revise 1st order DEs

First you need to write your equation in standard form by dividing by x, then work out the integrating factor by the standard method, then multiply by this IF and rewrite the equation as d/dx(something) = something else


(For the record, I get the IF to be $\dfrac{1}{x^2}$

Aha yes, I will definite go over it again!

But how did you get that as your IF? Isn't the formula to find the integrating factor e^integratePx

Your rewritten DE should be:

v' - (2v/x) = -1

so P = -2/x, integral of P = -2lnx = -(ln(x^2)) = ln(1/x^2). hence IF = 1/x^2.

Ah now I see! Thanks!

But just go back to our first few steps, when we have



how did we know that a) v wasn't a constant b) that after differentiating it would make dv/dx and not dv/dy

a) You don't (although this would only happen if y was very simple!)

b) It depends on what you differentiate with respect to. This is like your thread the other day - x is the independent variable, y is the original dependent variable, and v is another dependent variable defined by v = dy/dx. Your original DE involves d^2y/dx^2 so the obvious thing to do is to differentiate with respect to x.

a) what do you mean?

b) right, i see!

c) in the end i got the answer v/x^2 = -lnx +c which after i played around with gave me v=x^2lnC/x I wanted to convert the v into y but v=dy/dx

The answer I should get in the end is y = Ax^3 + 1/2x^2 + B

a) if v = dy/dx, v can only be constant if y = kx for some constant k!

c) I didn't get any logarithms in my working - don't forget you started with -x on the RHS and then divided everything by x to get -1 so it's a lot simpler than you've got.

I thought I did that

xv' - 2v = -x

so

v' - (2v/x) = -1

IF is 1/x^2 so:

(1/x^2)[v' - (2v/x)] = -1/x^2

or d/dx(v/x^2) = -1/x^2

Can you finish it from here?

I've got up to v/x^2 = 1/x + c