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Math Help - First Order Differential Equations: Solving Particular and Geneal Solution

  1. #1
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    First Order Differential Equations: Solving Particular and Geneal Solution

    How to do you solve these?

    1) find an explicit general solution of the differential equation:
    dy/dx= ysinx

    2) Find an implicit general solution of the differential equation:
    y^(3) dy/dx= (y^(4)+1) cosx

    3) Find an explicit particular solution of the initial value problem:
    x dy/dx - y= 2x^(2) y, y(1)= 1

    4) Find an explicit particular solution of the initial value problem:
    dy/dx= 6e^(2x-y), y(0)= 0

    5) A pitcher of buttermilk initially at 25 Degree celcius is to be cooled by setting it on the front porch, where the temperature is 0 degree celcius. Suppose that the temperature of the buttermilk has dropped to 15 Degree Celcius after 20 minutes, when will it be 5 degree celcius?
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  2. #2
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    Quote Originally Posted by googoogaga View Post
    1) find an explicit general solution of the differential equation:
    dy/dx= ysinx
    This one's easy. ("Divide and conquer." )

    \frac{dy}{dx} = y sin(x)

    \frac{dy}{y} = sin(x) dx

    \int \frac{dy}{y} = \int sin(x) dx

    ln(|y|) = -cos(x) + C

    |y| = e^{-cos(x) + C}

    or perhaps you would prefer the form:
    |y| = e^{-cos(x)}e^{C} = Ae^{-cos(x)}

    In either form the exponential is always positive (note then that we need to require that the constant A is positive) so we may drop the absolute value bars:
    y = Ae^{-cos(x)}

    -Dan
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by googoogaga View Post
    2) Find an implicit general solution of the differential equation:
    y^(3) dy/dx= (y^(4)+1) cosx
    This is the same idea as the first.

    y^3 \frac{dy}{dx} = (y^4 + 1)cos(x)

    \frac{y^3}{y^4 + 1} dy = cos(x) dx

    \int \frac{y^3}{y^4 + 1} dy = \int cos(x) dx

    For the LHS we may substitute z = y^4 + 1 ==> dz = 4y^3 dy, so

    \int \frac{y^3}{y^4 + 1} dy = \frac{1}{4} \int \frac{1}{z} = \int cos(x) dx

    \frac{1}{4}ln(|z|) = sin(x) + C

    |z| = Ae^{4sin(x)}<-- Just for the record, A = e^{4C}

    Again, we may drop the absolute value bars, but we must require that z = y^4 + 1 never be negative. But this is true anyway, so no worries!

    y^4 + 1 = Ae^{sin(x)}

    Since all the question asked for is an implicit relationship, you can be done with this line here.

    -Dan
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    Quote Originally Posted by googoogaga View Post
    3) Find an explicit particular solution of the initial value problem:
    x dy/dx - y= 2x^(2) y, y(1)= 1
    x \frac{dy}{dx} - y = 2x^2

    I forget whose name is attached to this equation, but we may simplify it somewhat by the substitution
    x = e^t

    which implies
    dx = e^t dt ==> \frac{d}{dx} = e^{-t}\frac{d}{dt}

    So the differential equation becomes:
    e^t \left ( e^{-t} \frac{dy}{dt} \right ) - y = 2e^{2t}

    \frac{dy}{dt} - y = 2e^{2t}

    The characteristic equation for the homogeneous equation is
    m - 1 = 0

    Thus m = 1 ==> y_h(t) = Ae^{t}

    The particular solution is provided by a term of the form y_p(t) = Be^{ct}. Substitution in the t version of the differential equation gives:
    Bce^{ct} - Be^{ct} = 2e^{2t}

    Thus c = 2 (by matching the exponents) and Bc - B = 2 (by matching the coefficients) thus B = 2 also.

    So the solution to the t version of the differential equation is
    y(t) = Ae^{t} + 2e^{2t}

    Now we need to revert back to the x variable. x = ln(t), so
    y(x) = Ae^{ln(x)} + 2e^{2ln(x)} = Ax + 2x^2

    Now we have the initial condition: y(1) = 1.
    y(1) = A(1) + 2(1)^2 = 2 + A = 1

    Thus A = -1

    and the solution to the differential equation will be:
    y(x) = 2x^2 - x

    -Dan
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by googoogaga View Post
    4) Find an explicit particular solution of the initial value problem:
    dy/dx= 6e^(2x-y), y(0)= 0
    This is done in the same manner as 1) and 2). I get that
    y(x) = ln(3e^{2x} - 2)

    Quote Originally Posted by googoogaga View Post
    5) A pitcher of buttermilk initially at 25 Degree celcius is to be cooled by setting it on the front porch, where the temperature is 0 degree celcius. Suppose that the temperature of the buttermilk has dropped to 15 Degree Celcius after 20 minutes, when will it be 5 degree celcius?
    Here we need to assume that the temperature change is a linear function (according to Newton's Law of Cooling it should be), so we have
    \frac{dT}{dt} = \text{constant} = \frac{25 ^oC - 15 ^oC}{-20~min} = -0.5 ^oC/min

    So the temperature starts as 25 ^oC and cools at a constant 0.5 ^oC/min. How long will it take to drop to 5 ^oC?

    -Dan
    Last edited by topsquark; July 11th 2007 at 05:45 PM.
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  6. #6
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    First Order Differential Equations: Solving Particual and general Solution...

    # 4 still isn't crystal clear. How do you get the time? I am not really sure what to do here. Could you further explain please? Thanks.
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  7. #7
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    Quote Originally Posted by googoogaga View Post
    4) Find an explicit particular solution of the initial value problem:
    dy/dx= 6e^(2x-y), y(0)= 0
    \frac{dy}{dx} = 6e^{2x - y} = 6e^{2x}e^{-y}

    e^y dy = 6e^{2x} dx

    \int e^y dy = \int 6e^{2x} dx

    Can you take it from here?

    -Dan
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  8. #8
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    Quote Originally Posted by googoogaga View Post
    How do you get the time? I am not really sure what to do here. Could you further explain please? Thanks.
    I presume you mean for #5. There's not much else I can say without just giving you the answer (and this one's too easy for that.)

    Let me put it this way:
    You are looking at a plane of graph paper where time, t, is on the x-axis and temperature, T, is on the y-axis. We have two points on the function we wish to graph:
    (0~min, 25 ^oC)
    and
    (20~min, 15 ^oC)

    I told you the function connecting these points is linear, that is to say, a line. I found the slope of this line, -0.5 ^oC/min. Now you need to supply the following coordinate: What is the t value when the T value is 5 ^oC? ie. Fill in the following t coordinate: (t, 5 ^oC).

    If you were asked this about the xy plane and a line connecting the points I am sure you would have no problem doing this.

    I should note that in writing this I realized that \frac{dT}{dt} is negative. I have fixed this in my original post on this question. Sorry for the error!

    -Dan
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