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Math Help - Linear Problem - Slope of Curve Through Point

  1. #1
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    Linear Problem - Slope of Curve Through Point

    Hi all,
    I have started on differential equations today and need some help (as usual).

    From what i have read so far.

    The solution to a differential equation will be another function or differential equation.

    For example the slope of a curve is a differential equation. The solution of this differential equation will actually give you the curve of the function or the equation of the function itself.

    Now i have this problem to solve.

    "find the curve whose slope at any point (x,y) is equal to 5y and which passes through the point (1,-2)".

    Now the slope is given as 5y;

     slope=\frac{dy}{dx}=5y

    Now this is a differential equation in itself and the solution to the differential equation will give me the curve of the function itself.

    rearranging the slope term

    multiply across by dx:

     \frac{dy}{dx}.dx=5y.dx

     dy=5y.dx

    To solve the differential equation i need to integrate both sides;

     \int dy=\int 5y.dx

     x+C1= \frac{5y^2}{2} +C2

    rearranging similar terms:

     x- \frac{5y^2}{2} =C2-C1

     x- \frac{5y^2}{2} =C.........equation (1)

    Now we are given the point (1,-2) so substitute into equation (1) for x and y

     (1)- \frac{5(-2)^2}{2} =C.........equation (1)

     C=-9

    Now the solution to the problem is C inserted into equation (1)

     x- \frac{5y^2}{2} =-9

    or

     x- \frac{5y^2}{2} +9=0

    Now to check this if i substitute the solution back into the original equation for y and if i differentiate y with respect to x i should get the slope of 5y?

    I think i understand it-some of the text overcomplicates this if you ask me(or more probably i have not got the brains to understand it),
    can anybody give me a more basic explanation if i have not got it correct,
    Is my method/solution correct?
    Thanks
    John
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  2. #2
    MHF Contributor

    Joined
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    re: Linear Problem - Slope of Curve Through Point

    Quote Originally Posted by celtic1234 View Post
    Hi all,
    I have started on differential equations today and need some help (as usual).

    From what i have read so far.

    The solution to a differential equation will be another function or differential equation.

    For example the slope of a curve is a differential equation. The solution of this differential equation will actually give you the curve of the function or the equation of the function itself.

    Now i have this problem to solve.

    "find the curve whose slope at any point (x,y) is equal to 5y and which passes through the point (1,-2)".

    Now the slope is given as 5y;

     slope=\frac{dy}{dx}=5y

    Now this is a differential equation in itself and the solution to the differential equation will give me the curve of the function itself.

    rearranging the slope term

    multiply across by dx:

     \frac{dy}{dx}.dx=5y.dx

     dy=5y.dx

    To solve the differential equation i need to integrate both sides;

     \int dy=\int 5y.dx

     x+C1= \frac{5y^2}{2} +C2
    NO!! \int 5y dy= 5y^2/2+ C but you cannot integrate y with respect to x! y is an unknown function of x.

    Instead, you need to separate the x and y terms:
    \frac{dy}{y}= 5 dx

    Now integrate
    \int \frac{dy}{y}= \int 5 dx

    rearranging similar terms:

     x- \frac{5y^2}{2} =C2-C1

     x- \frac{5y^2}{2} =C.........equation (1)

    Now we are given the point (1,-2) so substitute into equation (1) for x and y

     (1)- \frac{5(-2)^2}{2} =C.........equation (1)

     C=-9

    Now the solution to the problem is C inserted into equation (1)

     x- \frac{5y^2}{2} =-9

    or

     x- \frac{5y^2}{2} +9=0

    Now to check this if i substitute the solution back into the original equation for y and if i differentiate y with respect to x i should get the slope of 5y?

    I think i understand it-some of the text overcomplicates this if you ask me(or more probably i have not got the brains to understand it),
    can anybody give me a more basic explanation if i have not got it correct,
    Is my method/solution correct?
    Thanks
    John
    No, your method is not correct. In particular, you cannot integrate y, an unknown function of x, with respect to x. \int y dx is NOT the same as \int y dy.
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  3. #3
    Member
    Joined
    Jul 2011
    Posts
    80

    Re: Linear Problem - Slope of Curve Through Point

    ok,
    so if i am getting what you re saying;

     \frac{1}{y}.dy=5.dx

    integrate both sides:

     \int \frac{1}{dy}.dy=\int 5.dx

     ln y + C1= 5x + C2

     ln y-5x=C


    At point (1,-2)

     -ln (-2)-5(1)=C

     C =-5.69315

    Substituting into the solution for C

     ln y-5x=-5.69315

     ln y-5x+5.69315 =0

    Is this it?
    John
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