Results 1 to 4 of 4

Math Help - Seperable DE

  1. #1
    Member
    Joined
    Jan 2008
    Posts
    91

    Seperable DE

    Hey,

    I've been having some trouble recreating the answer my teacher has for the following question:

    Find y(x) given \frac{dy}{dx} = x^3y, where y(1) = 2.

    So I did the following:

    \int \frac{dy}{y} = \int x^3dx \longrightarrow \ln{y} = \frac{x^4}{4} + C

    y = e^{\frac{x^4}{4} + C} \ \ or \ \ y = Ce^{\frac{x^4}{4}}

    Where C is some constant to be solved for.

    Anyway I could solve for C. But what is getting me is that my prof's final answer is:

    y = 2e^{\frac{x^4}{4} - \frac{1}{4}}

    Which would mean there was two constants...but I would need more initial conditions for that.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641

    No

    Quote Originally Posted by TrevorP View Post
    Hey,

    I've been having some trouble recreating the answer my teacher has for the following question:

    Find y(x) given \frac{dy}{dx} = x^3y, where y(1) = 2.

    So I did the following:

    \int \frac{dy}{y} = \int x^3dx \longrightarrow \ln{y} = \frac{x^4}{4} + C

    y = e^{\frac{x^4}{4} + C} \ \ or \ \ y = Ce^{\frac{x^4}{4}}

    Where C is some constant to be solved for.

    Anyway I could solve for C. But what is getting me is that my prof's final answer is:

    y = 2e^{\frac{x^4}{4} - \frac{1}{4}}

    Which would mean there was two constants...but I would need more initial conditions for that.
    You have that 2=Ce^{\frac{1^4}{4}} dividing each side we get c=\frac{2}{e^{\frac{1}{4}}}\Rightarrow2e^{\frac{-1}{4}} see it now?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,867
    Thanks
    745
    Hello, Trevor!

    You're both correct . . .


    y \:= \:e^{\frac{x^4}{4} + C}\:\text{ or } \:y \:= \:Ce^{\frac{x^4}{4}} .where C is some constant to be solved for.

    I could solve for C. . . . . Did you?

    \text{My prof's final answer is: }\;y \:= \:2\,e^{\frac{x^4}{4} - \frac{1}{4}}

    Since y(1) = 2, we have: . Ce^{\frac{1}{4}} \:=\:2\quad\Rightarrow\quad C \:=\:2e^{\text{-}\frac{1}{4}}


    Then the function becomes: . y \;=\;2\underbrace{e^{\text{-}\frac{1}{4}}\cdot e^{\frac{x^2}{4}}}_{\text{combine}} \;=\;2\,e^{ \frac{x^2}{4}-\frac{1}{4}} . . . see?

    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Jan 2008
    Posts
    91
    Yeah I did..but something just wasn't clicking for some reason. I just thought that since two numbers were introduced there should be two variables.

    But now it makes sense.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Seperable DE Help!!
    Posted in the Differential Equations Forum
    Replies: 3
    Last Post: February 23rd 2010, 01:24 PM
  2. Help with a seperable variable? I think?
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: November 17th 2009, 04:29 PM
  3. [SOLVED] Seperable DE
    Posted in the Differential Equations Forum
    Replies: 4
    Last Post: June 3rd 2009, 03:04 PM
  4. Another Seperable Differential EQ
    Posted in the Calculus Forum
    Replies: 4
    Last Post: April 9th 2009, 03:07 PM
  5. Seperable Differential EQ
    Posted in the Calculus Forum
    Replies: 2
    Last Post: April 9th 2009, 12:34 PM

Search Tags


/mathhelpforum @mathhelpforum