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Homework Help: Separating variables and then finding their indefinite integrals

  1. Oct 1, 2011 #1
    Hi I have not studied calculus for a while and I am just seeking some clarification on the following two problems I have attempted to solve.


    dy/dx = y(x^3 - √x)

    I have separated the variables as follows:

    Rewrote equation as dy/dx = y(x^3 - x^1/2)

    Divided both sides by 1/y

    1/y dy/dx = x^3 - x^1/2

    Multiplied both sides by dx

    1/y dy = x^3 - x^1/2 dx

    Then to find the indefinite integral

    ∫1/y dy = ∫x^3 - x^1/2 dx

    ln[y] = 0.25x^4 - 1.5x^3/2

    y = e^(0.25x^4 - 1.5x^3/2) + C

    Does this look correct?


    (yx^4 - yx^2) = dy/dx y^2.x^4

    Separating variables:

    Multiply both sides by dx

    (yx^4 - yx^2) dx = y^2.x^4 dy

    Divide both sides by y

    x^4 - x^2 dx = y.x^4 dy

    Divide boths sides by x^4

    (x^4 - x^2)/x^4 dx = y dy

    Swap LHS and RHS so

    y dy = (x^4 - x^2)/x^4 dx

    Then to find the indefinite integral

    ∫y dy = ∫(x^4 - x^2)/x^4 dx

    y^2/2 = (x^2 + 1)/x + C

    Solve for y

    y = √2(x^2 + 1)/x + C

    If my solutions are incorrect then any pointers in the right direction would be gratefully received.

  2. jcsd
  3. Oct 1, 2011 #2
    The last term in the first problem is incorrect. Haven't checked the second problem yet.

    To check results, take the derivative of your answer and see if you can work backwards and get the original ODE.
  4. Oct 1, 2011 #3
    ((x**2+1)/x) should be under radical. Not sure of your notation.
  5. Oct 1, 2011 #4
    Will have another look at first problem. Is it just my transformation to separate y from ln [y] where I have gone wrong?

    The bracketed term in the second one is meant to be all under the radical.

    Thanks for your help.
  6. Oct 1, 2011 #5
    Look at the integral of the square root of x - first problem.
  7. Oct 1, 2011 #6
    Should the final answer be

    y = e^(0.25x^4 - 0.667x^3/2)
  8. Oct 1, 2011 #7
    Take the derivative with respect to x and see if you can get back to the original equation.
    d e^u = e^u * du

    My simply telling you does help you as much. You need to do it yourself.
  9. Oct 1, 2011 #8


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    Homework Helper

    It is more accurate to use 2/3 instead of the 0.667.

    The solution of the second problem is badly written. You must use parentheses.

    y^2/2 = (x^2 + 1)/x + C - this is correct

    y = √2(x^2 + 1)/x + C

    This is wrong, as the square root refers to 2 only. You have to write

    y=√[2(x^2 + 1)/x + C]

    or [itex]y=\sqrt{2\frac{x^2+1}{x}+C}[/itex].

    LawrenceC is right, always check your integral seeing if its derivative is equal to the original function.

  10. Oct 2, 2011 #9
    Thanks ehild.

    I had the correct layout of the solution to Problem 2 in my notes but just made an error typing it on here so it looked like I was only taking the square root of the 2.

    Will have another look at my solution to Problem 1.

  11. Oct 2, 2011 #10


    User Avatar
    Science Advisor

    Your Calculus is good- your algebra is awful!

    The constant of integration should be put in here:
    ln(y)= (1/4)x^4- (2/3)x^(3/2)+ C (and 1/(3/2)= 2/3, not 3/2)

    You cannot just arbitrarily add "C" here. C was the "constant of integration", add when you integrated. Taking the exponential of that means taking exponential of the whole thing:
    [tex]y= e^{(1/4)x^4- (2/3)x^{3/2}+ C}= C'e^{(1/4)x^4- (2/3)x^{3/2}}[/tex]
    where [itex]C'= e^C[/itex]. Since C is an "arbitrary constant", so is C' but it is multiplied by the solution, not added to it.

    Ambiguous? Where does the square root end? You should have
    y = √[2(x^2 + 1)/x + C]
    (strictly speaking it should be "2C" but, again, C is an arbitrary constant so 2C is also and it doesn't hurt to just write "C".

    Also it should be "plus or minus:
    [tex]y= \pm\sqrt{[2(x^2+ 1)/x+ C]}[/tex]

    Personally, I would have prefered leaving it as
    [tex]\frac{y^2}{2}= \frac{x^2+ 1}{x}+ C[/tex]
    [tex]y^2= \frac{2(x^2+ 1)}{x}+ C[/tex]

    [/quote]If my solutions are incorrect then any pointers in the right direction would be gratefully received.

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