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Math Help - 2 problems

  1. #1
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    2 problems

    Hi ,
    I have two problem with which I am stuck:

    1- Given the family of curves y=1/(x+C).
    Find the family of orthogonal trajectories.

    For this problem, I took the first derivative of y---> y'=-1/(x+C)^2. from here, I cannot find the value of C since (x+C)^2=-1/y' which is impossible.
    What can I do here?

    2- The pairs {x^2+x,x^2} and {x,x+2x^2} ( found by the Wronskian) are base solutions of the equation y''+p(x)y'+q(x)y=0.
    Give the general solution of the equation.

    Here I dont even know how to start.

    Thank you
    B
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by braddy
    Hi ,
    I have two problem with which I am stuck:

    1- Given the family of curves y=1/(x+C).
    Find the family of orthogonal trajectories.

    For this problem, I took the first derivative of y---> y'=-1/(x+C)^2. from here, I cannot find the value of C since (x+C)^2=-1/y' which is impossible.
    What can I do here?
    Given the family of curves

    <br />
y=\frac{1}{x+C}<br />
,

    rearrange to give:

    <br />
x+C=1/y<br />

    and differentiate:

    <br />
1=-\frac{1}{y^2}\ \frac{dy}{dx}<br />

    or:

    <br />
\frac{dy}{dx}=-y^2<br />
.

    So now we know that the slope of any curve of the family that passes
    through (x,y) is -y^2. So the slope of a curve orthogonal to a member
    of the family passing through (x,y) is 1/y^2. So a curve in the orthoganal family
    satisfies the ODE:

    <br />
\frac{dy}{dx}=\frac{1}{y^2}<br />

    Which has solutions:

    <br />
f(x,y)=\frac{y^3}{3}-x+K=0<br />

    Which is the required orthogonal family.

    RonL
    Last edited by CaptainBlack; February 25th 2006 at 11:35 PM.
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by braddy

    2- The pairs {x^2+x,x^2} and {x,x+2x^2} ( found by the Wronskian) are base solutions of the equation y''+p(x)y'+q(x)y=0.
    Give the general solution of the equation.

    Here I dont even know how to start.
    Now you have me slightly confused here. As I understand things a pair
    \{f(x),g(x)\} is a base solution of y''+p(x)y'+q(x)y=0
    means that f and g are linearly independent
    (have non-zero Wronskian), and Af(x)+Bg(x) is a general
    solution of y''+p(x)y'+q(x)y=0.

    Your pairs both can be used to give a general solution, and with a bit
    of rearrangement can be seen to give the same solutions for suitable
    choices of the multipliers.

    RonL
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  4. #4
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    Quote Originally Posted by CaptainBlack
    Now you have me slightly confused here. As I understand things a pair
    \{f(x),g(x)\} is a base solution of y''+p(x)y'+q(x)y=0
    means that f and g are linearly independent
    (have non-zero Wronskian), and Af(x)+Bg(x) is a general
    solution of y''+p(x)y'+q(x)y=0.

    Your pairs both can be used to give a general solution, and with a bit
    of rearrangement can be seen to give the same solutions for suitable
    choices of the multipliers.

    RonL
    Yes you are right,
    if you substract the two function that contitute the basis, I have the same general solution fot both sets:
    Y=C1*x + C2 X^2.

    thank you for both solutions.
    B
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