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Math Help - First Order Differential Equations: Separable Equations and Applications

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    First Order Differential Equations: Separable Equations and Applications

    I please need your help with the following homework problems. I want to make sure that I followed the correct procedure in solving these problems. Thanks!

    1. Solve the Initial Value Problem:
    a) dy/dx= (x)^1/2, y(4)= 0
    b) dy/dx= cos 2x, y(0)= 1
    c) dy/dx= xe^(-x), y(0)=1

    2. A moving particle has acceleration a(t)=5, intial velocity v=1, and initial position x=2. Find the position function x(t) of the particle.

    3. A moving particle has acceleration a(t)= 2t+1, initial velocity v=-7, and initial position x=4. Find the position function x(t) of the particle.

    4. The skid marks made by an automoblie indicated that its brakes were fully applied for a distance of 75m before it came to a stop. The car in quesion is known to have a constant deceleration of 20 m/s^2 under these conditions. How fast - in km/h - was the car traveling when the brakes were first applied?
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    Quote Originally Posted by googoogaga View Post
    I please need your help with the following homework problems. I want to make sure that I followed the correct procedure in solving these problems. Thanks!

    1. Solve the Initial Value Problem:
    a) dy/dx= (x)^1/2, y(4)= 0
    b) dy/dx= cos 2x, y(0)= 1
    c) dy/dx= xe^(-x), y(0)=1
    since you are only concerned about procedure, and all parts of 1 are similar, i will only do (a). If concerned, you can tell us your solutions for the others and we'll tell you if they're right or not.

    (a)

    \frac {dy}{dx} = x^{1/2}

    \Rightarrow y = \int x^{1/2}~dx

    \Rightarrow y = \frac {2}{3} x^{3/2} + C

    Now we are told that y(4) = 0 (that is, when x = 4, y = 0):

    \Rightarrow 0 = \frac {2}{3} 4^{3/2} + C

    \Rightarrow C = - \frac {16}{3}

    \implies \boxed { y = \frac {2}{3} x^{3/2} - \frac {16}{3} }
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by googoogaga View Post

    2. A moving particle has acceleration a(t)=5, intial velocity v=1, and initial position x=2. Find the position function x(t) of the particle.

    3. A moving particle has acceleration a(t)= 2t+1, initial velocity v=-7, and initial position x=4. Find the position function x(t) of the particle.
    i will only do 2, since 3 is similar

    2.
    You should have some background knowledge here. You are expected to know that.

    (we usually use s, but since you use x, i will also)

    if x(t) = position function

    v(t) = x'(t) = velocity function

    a(t) = v'(t) = acceleration function

    So we have:

    a(t) = 5

    \Rightarrow v(t) = \int a(t)~dt = \int 5 ~dt = 5t + C

    we are told that v(0) = 1 (that is, when t = 0, v = 1):

    \Rightarrow 1 = 5(0) + C \implies C = 1

    Thus \boxed { v(t) = 5t + 1 }

    \Rightarrow x(t) = \int v(t)~dt = \int (5t + 1)~dt = \frac {5}{2} t^2 + t + D

    we are told that x(0) = 2 (that is, when t = 0, x = 2):

    \Rightarrow 2 = D

    \Rightarrow \boxed { x(t) = \frac {5}{2} t^2 + t + 2 }


    You could also use SUVAT equations to answer these questions, but i figured this was the method you were after. The SUVAT equations were derived from similar procedures I would think
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by googoogaga View Post
    4. The skid marks made by an automoblie indicated that its brakes were fully applied for a distance of 75m before it came to a stop. The car in quesion is known to have a constant deceleration of 20 m/s^2 under these conditions. How fast - in km/h - was the car traveling when the brakes were first applied?
    use SUVAT equations for this

    Let s be displacement
    Let u be initial velocity
    Let v be final velocity
    Let a be acceleration
    Let t be time

    Our knowns are:
    a = -20 m/s^2

    s = 75 m

    v = 0 m/s

    Remember to watch your units, you can work in meters and seconds and convert at the end, or convert and work in the right units from the beginning

    Use v^2 = u^2 + 2as to solve for u

    And that's it
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