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Homework Help: Second-Order ODE with Missing X

  1. Oct 6, 2014 #1
    1. The problem statement, all variables and given/known data
    Solve the given differential equations with missing x.
    [itex] y'' + y = 0 [/itex]
    2. Relevant equations
    [itex] y = c_1cos(x) + c_2sin(x) [/itex]
    This is the answer given in the back of the book. However, I can't sem to get my answer to agree
    3. The attempt at a solution
    First, I made some substitutions:
    [itex] y' = v y'' = v' [/itex]
    Using the chain rule I obtained:
    [itex] \frac{dv}{dx} = v\frac{dv}{dy} = v' [/itex]
    So the equation becomes:
    [itex] v\frac{dv}{dy} + y = 0 [/itex]
    [itex] vdv = -ydy [/itex]
    [itex] \int vdv = -\int ydy [/itex]
    [itex] \frac{v^2}{2} = - \frac{y^2}{2} + C_1 [/itex]
    [itex] v = +- sqrt(-y + C_1) [/itex]
    [itex] \frac{dy}{dx} = sqrt(-y + C_1) [/itex]
    [itex] \int\frac{dy}{dx} = \int sqrt(-y+C_1),dx [/itex]
    [itex] y = \frac{2}{3} (-y+C_1)^{\frac{3}{2}} + C_2 [/itex]
    Clearly this solution is nowhere close to the one in the back of the book, so I was hoping somebody could point out where I have gone wrong :)
  2. jcsd
  3. Oct 6, 2014 #2
    Just to be clear, at the top the two equations are y'=v and y''=v'...it is not all one equation
  4. Oct 6, 2014 #3

    Ray Vickson

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    You should always plug in a "solution" to see if it really works. Have you done that?

    Surely your textbook must deal with problems similar to this one. Have you checked?
  5. Oct 6, 2014 #4


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    This is correct.

    But this isn't. Multiplying the first equation by 2 gives [itex]v^2= -y^2+ 2C_1[/itex]
    I wouldn't complain about your having "[itex]C_1[/itex]" rather than "[itex]2C_2[/itex]" (they are both "undetermined constants") but the difference between "[itex]y[/itex]" and "[itex]y^2[/itex] is very important!

  6. Oct 6, 2014 #5
    I was quite certain my solution didn't work, I was just unsure of where I was going wrong. Also, unfortunately my textbook does not provide any examples similar to this one, this textbook is pretty lousy and I'm considering buying a new one just so I don't run into problems like this...
  7. Oct 6, 2014 #6
    thank you
  8. Oct 6, 2014 #7
    ok, so after adjusting my answer:
    [itex] v = sqrt(-y^2 + C_1) [/itex]
    [itex] \int \frac{dy}{dx} = \int sqrt(-y^2 + C_1) dx [/itex]
    [itex] y = \frac{1}{3} (-y^2 + C_1)^{\frac{3}{2}} + C_2 [/itex]
    I am confused about how to go from the above answer to:
    [itex] C_1cos(x) + C_2sin(x) [/itex]
  9. Oct 6, 2014 #8

    Ray Vickson

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    OK, but never forget that Google is your friend. Looking at "linear differential equation" will turn up a host of relevant articles.

    Anyway, to verify that the book's "solution" is valid involves nothing more than plugging it in to see if it works. Of course, figuring out how the book got its solution in the first place is another issue. Basically, though, you just 'learn' that certain types of de's have certain types of solutions--- and that has been part of the math toolkit for more than 150 years.
  10. Oct 6, 2014 #9
    thank you, and while we're at it would you happen to know off the top of your head of any good DiffEq textbooks? Not only is my book short on examples but it is also chock full of small errors that can be very confusing at times :(
  11. Oct 6, 2014 #10


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    @n3wton: Look in your text for "constant coefficient" differential equations. Even a bad text will have that topic. Otherwise Google it.
  12. Oct 6, 2014 #11
    thank you, i'll search for it
  13. Oct 8, 2014 #12


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    The last line is wrong. You can not integrate like that.

  14. Oct 8, 2014 #13


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    This seems a roundabout way to solve this DE. Have you studied how to form the characteristic equation for an ODE yet?
  15. Oct 8, 2014 #14
    The problem gave specific orders that we were to solve it using this method. However, we have not yet studied what you mentioned...
  16. Oct 8, 2014 #15
    no? I did:
    [itex] u = -y^2 + C_1 [/itex]
    [itex] du = -2y dy [/itex]
    [itex] = \frac{-1}{2} \int -2(-y^{2} + C_1) dy [/itex]
    [itex] = \frac{-1}{2} \int u^{\frac{1}{2}} du [/itex]
    [itex] = \frac{-1}{2} \frac{2}{3} u^{\frac{3}{2}} [/itex]
    [itex] = \frac{-1}{3} (-y^2 + C_1)^{\frac{3}{2}} + C_2 [/itex]
    what is wrong with that?
  17. Oct 8, 2014 #16


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    Everything. The left side is a differential, the right side is an integral. You mixed u and y. .. Do not forget, you need to find y(x).

  18. Oct 8, 2014 #17
    Ok, I'm confused now, I thought since v = dy/dv I could make that substitution and then integrate, would I first need to sub x for y in the equation to be integrated?
  19. Oct 8, 2014 #18


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    You can integrate a function of x with respect to x and a function of y with respect to y.

    You have ##\frac{dy}{dx} =sqrt(-y^2 + C_1) ##

    Collect the y terms on one side and the x terms on the other side.


    and integrate: [tex]\int{\frac{dy}{\sqrt{-y^2+C_1}}}=\int{dx}[/tex]

    You get something like ##-\arccos( y/\sqrt{C_1})=x+C_2##
    Solve for y.

    Last edited: Oct 8, 2014
  20. Oct 9, 2014 #19
    Thank you so much! The book's solution makes soo much more sense now...I'm on my phone now but I'll try to formulate a final solution once I'm off campus :)
  21. Oct 9, 2014 #20
    Ok, I am getting something slightly different for my integral, but my trig sub skills are a bit rusty, here is what I did:
    [itex] \int \frac{dy}{sqrt(C_1-y^2)} [/itex]
    [itex] a = sqrt(C_1) [/itex]
    [itex] y = sqrt(C_1)Sinb [/itex]
    [itex] sqrt(C_1 - y^2) = sqrt(C_1)cosb [/itex]
    [itex] dy = sqrt(C_1)Cosb [/itex]
    [itex] = \int \frac{sqrt(C_1)Cosb db}{sqrt(C_1)Cosb} [/itex]
    [itex] = \int db [/itex]
    [itex] b = arccos(\frac{sqrt(C_1 - y^2)}{sqrt(C_1))} [/itex]
    So then I would solve for Y in this equation:
    [itex] arccos('') = x + C_2 [/itex] ?
    Last edited: Oct 9, 2014
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