# Thread: First Order Differential Equations: Separable Equations and Applications

1. ## 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?

2. Originally Posted by googoogaga 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)

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

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

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

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

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

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

$\displaystyle \implies \boxed { y = \frac {2}{3} x^{3/2} - \frac {16}{3} }$

3. Originally Posted by googoogaga 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:

$\displaystyle a(t) = 5$

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

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

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

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

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

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

$\displaystyle \Rightarrow 2 = D$

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

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

4. Originally Posted by googoogaga 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 $\displaystyle s$ be displacement
Let $\displaystyle u$ be initial velocity
Let $\displaystyle v$ be final velocity
Let $\displaystyle a$ be acceleration
Let $\displaystyle t$ be time

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

$\displaystyle s = 75$ m

$\displaystyle v = 0$ 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 $\displaystyle v^2 = u^2 + 2as$ to solve for $\displaystyle u$

And that's it

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