Second derivative differential equations in terms of y?

  • #1
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Firstly I know how to do this with first derivatives in differential equations - for example say we had ##\frac{dy}{dx}=4y^2-y##, and we're also told that ##y=1## when ##x=0##.

##\frac{dy}{dx}=4y^2-y##
##\frac{dx}{dy}=\frac{1}{4y^2-y}=\frac{1}{y\left(4y-1\right)}=\frac{4}{4y-1}-\frac{1}{y}##
##\int _{ }^{ }\frac{dx}{dy}dy=\int _{ }^{ }\frac{4}{4y-1}dy-\int _{ }^{ }\frac{1}{y}dy##
##x=\ln \left(\left|4y-1\right|\right)-\ln \left(\left|y\right|\right)+c##

Substituting in ##y=1## and ##x=0##, we can find ##c##:
##x=\ln \left(\left|4y-1\right|\right)-\ln \left(\left|y\right|\right)+c##
##0=\ln \left(3\right)+c,\ c=-\ln \left(3\right)##
##x=\ln \left(\left|4y-1\right|\right)-\ln \left(\left|y\right|\right)-\ln \left(3\right)##
##x=\ln \left(\frac{\left|4y-1\right|}{3\left|y\right|}\right)##
##3e^x=\left|4-\frac{1}{y}\right|##
##4-\frac{1}{y}=-3e^x,4-\frac{1}{y}=3e^x##

Because we know that ##y## has to equal postive ##1## when ##x## equals ##0##;
##4-1=-3e^0,\ 4-1=3e^0##
##3=-3,\ 3=3##
As the first statement is false, the second one must be true.
##∴4-\frac{1}{y}=3e^x##
##\frac{1}{y}=4-3e^x##
##y=\frac{1}{4-3e^x}##.

So as you can see it's relatively simple to solve a differential equation of the form ##\frac{dy}{dx}=f\left(y\right)##, but I'm wondering if it would be possible to solve one in the form of ##\frac{d^2y}{dx^2}=f\left(y\right)##. I've tried looking it up online but I haven't been able to find anything useful.
 

Answers and Replies

  • #2
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So as you can see it's relatively simple to solve a differential equation of the form ##\frac{dy}{dx}=f\left(y\right)##, but I'm wondering if it would be possible to solve one in the form of ##\frac{d^2y}{dx^2}=f\left(y\right)##. I've tried looking it up online but I haven't been able to find anything useful.
One can introduce ##z:=\frac{dy}{dx}=f\left(x,y\right)## and get a system of two differential equations of order one. The other equation is ##\frac{dz}{dx}=f\left(x,y,z\right)##. This process can be iterated to higher orders. You can look for ODE (ordinary differential equations) to solve such systems.
 
  • #3
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So as you can see it's relatively simple to solve a differential equation of the form ##\frac{dy}{dx}=f\left(y\right)##,
It's not necessarily that simple -- you have to be able to integrate ##\int \frac{dy}{f(y)}##.
 
  • #4
pasmith
Homework Helper
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So as you can see it's relatively simple to solve a differential equation of the form ##\frac{dy}{dx}=f\left(y\right)##, but I'm wondering if it would be possible to solve one in the form of ##\frac{d^2y}{dx^2}=f\left(y\right)##. I've tried looking it up online but I haven't been able to find anything useful.
Multiply both sides by [itex]y'[/itex] and integrate once with respect to [itex]x[/itex] to obtain [tex]
\left(\frac{dy}{dx}\right)^2 = 2\int f(y)\,dy.[/tex] Now just determine the constant of integration from the initial conditions, and work out whether you need the positive root or the negative root, and the problem is reduced to a first-order separable ODE.
 
  • #5
143
7
Multiply both sides by [itex]y'[/itex] and integrate once with respect to [itex]x[/itex] to obtain [tex]
\left(\frac{dy}{dx}\right)^2 = 2\int f(y)\,dy.[/tex] Now just determine the constant of integration from the initial conditions, and work out whether you need the positive root or the negative root, and the problem is reduced to a first-order separable ODE.
Thank you for you're help but I still find myself a little confused about how you arrived at [itex]\left(\frac{dy}{dx}\right)^2 = 2\int f(y)\,dy.[/itex]

I tried multiplying both sides by ##\frac{dy}{dx}## and then integrating both sides with respect to ##x## and this is what I got;
[tex]\frac{d^2y}{dx^2}=f\left(y\right)[/tex]
[tex]\frac{dy}{dx}\cdot \frac{d}{dx}\cdot \frac{dy}{dx}=f\left(y\right)\cdot \frac{dy}{dx}[/tex]
[tex]\left(\frac{dy}{dx}\right)^2\cdot \frac{d}{dx}=f\left(y\right)\cdot \frac{dy}{dx}[/tex]
[tex]\int _{ }^{ }\left(\frac{dy}{dx}\right)^2\cdot \frac{d}{dx}\cdot dx=\int _{ }^{ }f\left(y\right)\cdot \frac{dy}{dx}\cdot dx[/tex]
[tex]\int _{ }^{ }\left(\frac{dy}{dx}\right)^2\cdot d=\int _{ }^{ }f\left(y\right)\cdot dy[/tex]

I am unsure of what to do on the left hand side - how am I supposed to integrate without a variable for which to integrate with respect to. I'm also not sure why you have a two out the front of the integral on the right hand side, but I'm guessing that the answer to my first question will probably explain that as well.
 

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