Differential Equations Application Problems

• Sep 10th 2010, 03:53 AM
Paymemoney
Differential Equations Application Problems
Hi
Need help on a few questions:

1) The equation of motion of a mass (on the end of a vertical spring) subject to an "input force" (which causes a forced motion) is

$\displaystyle \frac{d^2y}{dt^2} + 9y = 5cos(3t)$

Find y given that when t=0, y=0 and y' = 0

I get
$\displaystyle y = y_p +y_h$

$\displaystyle y = \frac{5}{6}tsin(3t) - \frac{5}{6}sin(3t)$

book's answer is $\displaystyle y= \frac{5}{6}tsin(3t)$

2) a) A particle moving in a straight line has displacement x, velocity v and acceleration a at time t, where $\displaystyle a=\frac{dv}{dt}=v\frac{dv}{dx}$. These formulas can be used to find v given a. The former is used when v is required in terms of t, the latter for v in terms of x.

b) A particle moves with retardation proportional to its velocity, ie, a = -kv, and has initial velocity $\displaystyle v_0$ and initial displacement 0. Find the velocity (i) in terms of time; and (ii) in terms of the displacement.

How would i do this question?

P.S
• Sep 10th 2010, 03:59 AM
mr fantastic
Quote:

Originally Posted by Paymemoney
Hi
Need help on a few questions:

1) The equation of motion of a mass (on the end of a vertical spring) subject to an "input force" (which causes a forced motion) is

$\displaystyle \frac{d^2y}{dt^2} + 9y = 5cos(3t)$

Find y given that when t=0, y=0 and y' = 0

I get
$\displaystyle y = y_p +y_h$

$\displaystyle y = \frac{5}{6}tsin(3t) - \frac{5}{6}sin(3t)$ Mr F says: Does this 'solution' satisfy y' = 0 ....? If you post your calculations, your mistake(s) can be pointed out.

book's answer is $\displaystyle y= \frac{5}{6}tsin(3t)$

2) a) A particle moving in a straight line has displacement x, velocity v and acceleration a at time t, where $\displaystyle a=\frac{dv}{dt}=v\frac{dv}{dx}$. These formulas can be used to find v given a. The former is used when v is required in terms of t, the latter for v in terms of x.

b) A particle moves with retardation proportional to its velocity, ie, a = -kv, and has initial velocity $\displaystyle v_0$ and initial displacement 0. Find the velocity (i) in terms of time; and (ii) in terms of the displacement.

How would i do this question?

P.S

2) b) (i) Solve dv/dt = -kv subject to $\displaystyle v(0) = v_0$.

2) b) (ii) Solve v dv/dx = -kv subject to $\displaystyle v = v_0$ when x = 0.
• Sep 10th 2010, 04:16 AM
Paymemoney
ok, heres my workings:

$\displaystyle y_h = Acos(3t)+Bsin(3t)$

$\displaystyle y_p = Atsin(3t)+Btcos(3t)$

$\displaystyle y_p' = 3tcos(3t)+Asin(3t)-3Btsin(3t)+Bcos(3t)$
$\displaystyle y_p'' = -9Atsin(3t)+3Acos(3t)+3Acos(3t)-9Btcos(3t)-3Bsin(3t)-3Bsin(3t)$
$\displaystyle y_p'' = -9Atsin(3t)+6Acos(3t)-9Btcos(3t)-6Bsin(3t)$

sub back into original equation:

$\displaystyle -9Atsin(3t)+6Acos(3t)-9Btcos(3t)-6Bsin(3t)+ 9Atsin(3t)+9Btcos(3t) = 5cos(3t)$
$\displaystyle 6Acos(3t)-6Bsin(3t)=5cos(3t)=5cos(3t)$
$\displaystyle 6A=5$
$\displaystyle A=\frac{5}{6}$

$\displaystyle -6B=0$
$\displaystyle B=0$

$\displaystyle y_p = \frac{5}{6}tsin(3t)$

t=0, y=0, y'=0

for y(0)=0
$\displaystyle 0=Acos0+Bsin0+\frac{5}{6}sin0$
$\displaystyle A=0$

$\displaystyle y'(0)=0$

$\displaystyle y'=-3Asin(3t)+3Bcos(3t)+\frac{15}{6}cos(3t)$
$\displaystyle 0=-3Asin(30)+3Bcos(30)+\frac{15}{6}cos(30)$
$\displaystyle 3B=\frac{-15}{6}$
$\displaystyle B=\frac{-15}{6}*\frac{1}{3}$

$\displaystyle y_h=\frac{-5}{6}sin(3t)$

$\displaystyle y=\frac{5}{6}tsin(3t)-\frac{5}{6}sin(3t)$
• Sep 10th 2010, 01:26 PM
mr fantastic
Quote:

Originally Posted by Paymemoney
ok, heres my workings:

$\displaystyle y_h = Acos(3t)+Bsin(3t)$

$\displaystyle y_p = Atsin(3t)+Btcos(3t)$

$\displaystyle y_p' = 3tcos(3t)+Asin(3t)-3Btsin(3t)+Bcos(3t)$
$\displaystyle y_p'' = -9Atsin(3t)+3Acos(3t)+3Acos(3t)-9Btcos(3t)-3Bsin(3t)-3Bsin(3t)$
$\displaystyle y_p'' = -9Atsin(3t)+6Acos(3t)-9Btcos(3t)-6Bsin(3t)$

sub back into original equation:

$\displaystyle -9Atsin(3t)+6Acos(3t)-9Btcos(3t)-6Bsin(3t)+ 9Atsin(3t)+9Btcos(3t) = 5cos(3t)$
$\displaystyle 6Acos(3t)-6Bsin(3t)=5cos(3t)=5cos(3t)$
$\displaystyle 6A=5$
$\displaystyle A=\frac{5}{6}$

$\displaystyle -6B=0$
$\displaystyle B=0$

$\displaystyle y_p = \frac{5}{6}tsin(3t)$

t=0, y=0, y'=0

for y(0)=0
$\displaystyle 0=Acos0+Bsin0+\frac{5}{6}sin0$
$\displaystyle A=0$

$\displaystyle y'(0)=0$

$\displaystyle y'=-3Asin(3t)+3Bcos(3t)+\frac{15}{6}cos(3t)$
$\displaystyle 0=-3Asin(30)+3Bcos(30)+\frac{15}{6}cos(30)$
$\displaystyle 3B=\frac{-15}{6}$
$\displaystyle B=\frac{-15}{6}*\frac{1}{3}$

$\displaystyle y_h=\frac{-5}{6}sin(3t)$

$\displaystyle y=\frac{5}{6}tsin(3t)-\frac{5}{6}sin(3t)$

Your calculation is good and will benefit many members. Sorry that I misread your original post. The book is wrong.
• Sep 10th 2010, 04:25 PM
Paymemoney
Quote:

Originally Posted by mr fantastic
2) b) (i) Solve dv/dt = -kv subject to $\displaystyle v(0) = v_0$.

this is what i got for i)
$\displaystyle \frac{dv}{dt}=-kv$

$\displaystyle v=\frac{-kv^2}{2}+C$
$\displaystyle v(0)=v_0$
$\displaystyle v_0=C$

therefore i get $\displaystyle \frac{-kv^2}{2}+v_0$
however this is not correct.
• Sep 11th 2010, 03:57 AM
mr fantastic
Quote:

Originally Posted by Paymemoney
this is what i got for i)
$\displaystyle \frac{dv}{dt}=-kv$

$\displaystyle v=\frac{-kv^2}{2}+C$
$\displaystyle v(0)=v_0$
$\displaystyle v_0=C$

therefore i get $\displaystyle \frac{-kv^2}{2}+v_0$
however this is not correct.

You have integrated the right hand side with respect to v and the left hand side with respect to t. You're expected to know the technique for this type of DE:

The next step $\displaystyle \frac{dt}{dv}=- \frac{1}{kv}$ and then you integrate both sides with respect to v ....