# Thread: Differential Equations Application Problems

1. ## 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

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

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

I get
$y = y_p +y_h$

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

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

2) a) A particle moving in a straight line has displacement x, velocity v and acceleration a at time t, where $a=\frac{dv}{dt}=v\frac{dv}{dx}$. 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 $v_0$ 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. 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

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

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

I get
$y = y_p +y_h$

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

2) a) A particle moving in a straight line has displacement x, velocity v and acceleration a at time t, where $a=\frac{dv}{dt}=v\frac{dv}{dx}$. 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 $v_0$ 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 $v(0) = v_0$.

2) b) (ii) Solve v dv/dx = -kv subject to $v = v_0$ when x = 0.

3. ok, heres my workings:

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

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

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

sub back into original equation:

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

$-6B=0$
$B=0$

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

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

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

$y'(0)=0$

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

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

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

4. Originally Posted by Paymemoney
ok, heres my workings:

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

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

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

sub back into original equation:

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

$-6B=0$
$B=0$

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

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

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

$y'(0)=0$

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

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

$y=\frac{5}{6}tsin(3t)-\frac{5}{6}sin(3t)$
Your calculation is good and will benefit many members. Sorry that I misread your original post. The book is wrong.

5. Originally Posted by mr fantastic
2) b) (i) Solve dv/dt = -kv subject to $v(0) = v_0$.
this is what i got for i)
$\frac{dv}{dt}=-kv$

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

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

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

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

therefore i get $\frac{-kv^2}{2}+v_0$
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 $\frac{dt}{dv}=- \frac{1}{kv}$ and then you integrate both sides with respect to v ....