Results 1 to 5 of 5

Math Help - Linear integral (still..)

  1. #1
    Member
    Joined
    May 2008
    Posts
    171

    Linear integral (still..)

    How am I suppose to solve this one?
    What is the way?
    Attached Thumbnails Attached Thumbnails Linear integral (still..)-1.jpg  
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Dec 2007
    From
    Anchorage, AK
    Posts
    276

    Convert to integral over x

    Sorry for the notation, but LaTeX appears to be mostly down.

    The endpoints of L are (0,1) and (t,y(t)). Now, we can use the fact that the curve L is given by y as a function y(x) of x to take:
    dy=(dy/dx)dx=y'(x)dx, and so our equation then becomes:
    ∫_{1}^{t}(-5x^3+3y)dx+xy'dx=(t^4-1)/4
    ∫_{1}^{t}(-5x^3+3y+xy')dx=(t^4-1)/4.
    (∫_{1}^{t} means definite integral from 1 to t).
    Now, we differentiate both sides with respect to t. Using the fundamental theory of calculus on the left hand side, we get:
    (d/dt)∫_{1}^{t}(-5x^3+3y+xy')dx=(d/dt)[(t^4-1)/4].
    -5t^3+3y(t)+ty'(t)=t^3
    ty'(t)+3y(t)=6t^3.
    Or, renaming the independent variable back to x,
    xy'(x)+3y(x)=6x^3.
    This is a nonhomogenous first-order linear differential equation, and should probably be solved by finding an integrating factor; in this case, we see it is x^2, as (d/dx)(x^3y)=x^3y'+3x^2y=x^2(xy'+3y),
    and so we have:
    x^3y'+3x^2y=6x^5
    (d/dx)(x^3y)=6x^5
    You can integrate with respect to x on both sides (don't forget the constant of integration), and then solve for y. Then find the value of the constant such that y(1)=0, and you have your solution.

    --Kevin C.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    May 2008
    Posts
    171
    Quote Originally Posted by TwistedOne151 View Post
    Sorry for the notation, but LaTeX appears to be mostly down.

    The endpoints of L are (0,1) and (t,y(t)). Now, we can use the fact that the curve L is given by y as a function y(x) of x to take:
    dy=(dy/dx)dx=y'(x)dx, and so our equation then becomes:
    ∫_{1}^{t}(-5x^3+3y)dx+xy'dx=(t^4-1)/4
    ∫_{1}^{t}(-5x^3+3y+xy')dx=(t^4-1)/4.
    (∫_{1}^{t} means definite integral from 1 to t).
    Now, we differentiate both sides with respect to t. Using the fundamental theory of calculus on the left hand side, we get:
    (d/dt)∫_{1}^{t}(-5x^3+3y+xy')dx=(d/dt)[(t^4-1)/4].
    -5t^3+3y(t)+ty'(t)=t^3
    ty'(t)+3y(t)=6t^3.
    Or, renaming the independent variable back to x,
    xy'(x)+3y(x)=6x^3.
    This is a nonhomogenous first-order linear differential equation, and should probably be solved by finding an integrating factor; in this case, we see it is x^2, as (d/dx)(x^3y)=x^3y'+3x^2y=x^2(xy'+3y),
    and so we have:
    x^3y'+3x^2y=6x^5
    (d/dx)(x^3y)=6x^5
    You can integrate with respect to x on both sides (don't forget the constant of integration), and then solve for y. Then find the value of the constant such that y(1)=0, and you have your solution.

    --Kevin C.

    TwistedOne151, I have a question about what you wrote here:
    (d/dt)∫_{1}^{t}(-5x^3+3y+xy')dx=(d/dt)[(t^4-1)/4].
    -5t^3+3y(t)+ty'(t)=t^3
    On the left side of the first equation you have an integral with respect to x, and you differentiate it with respect to t, and you come up with the same equation, only with t as the variable, how can you do that?

    10x again.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member
    Joined
    Dec 2007
    From
    Anchorage, AK
    Posts
    276

    Fundamental theorem of calculus

    You are familiar with the fundamental theorem of calculus, yes? I just used the form given here.
    We have the derivative with respect to t of the integral ∫_{1}^{t}(-5x^3+3y+xy')dx. Now, from the fundamental theorem of calculus, if F(x) is an antiderivative of the above (that is, if F'(x)=-5x^3+3y+xy'), then the integral is
    ∫_{1}^{t}(-5x^3+3y+xy')dx=F(t)-F(1),
    and so:
    (d/dt)∫_{1}^{t}(-5x^3+3y+xy')dx=(d/dt)(F(t)-F(1))
    =F'(t)-0
    =-5t^3+3y(t)+ty'(t)

    Does this explain it?

    --Kevin C.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    May 2008
    Posts
    171
    Quote Originally Posted by TwistedOne151 View Post
    You are familiar with the fundamental theorem of calculus, yes? I just used the form given here.
    We have the derivative with respect to t of the integral ∫_{1}^{t}(-5x^3+3y+xy')dx. Now, from the fundamental theorem of calculus, if F(x) is an antiderivative of the above (that is, if F'(x)=-5x^3+3y+xy'), then the integral is
    ∫_{1}^{t}(-5x^3+3y+xy')dx=F(t)-F(1),
    and so:
    (d/dt)∫_{1}^{t}(-5x^3+3y+xy')dx=(d/dt)(F(t)-F(1))
    =F'(t)-0
    =-5t^3+3y(t)+ty'(t)

    Does this explain it?

    --Kevin C.

    Yes, 10x
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 7
    Last Post: October 10th 2011, 03:06 PM
  2. 1st order linear equation with integral can not figure out
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: May 27th 2011, 10:40 AM
  3. Linear Algebra Integral
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: April 26th 2011, 02:01 AM
  4. Replies: 6
    Last Post: September 2nd 2009, 01:13 PM
  5. Linear integral
    Posted in the Calculus Forum
    Replies: 5
    Last Post: July 23rd 2008, 11:43 PM

Search Tags


/mathhelpforum @mathhelpforum