# Math Help - Two tank brine problem

1. ## Two tank brine problem

I am studying for my final and this is one of the problems in out trial final, ive gone through it about 5-6 times and i can not for the life of me get it right.

the pic and question is attached

What i have is

dx/dt=2-(3x/100)+(y/200)
dy/dt=(3x/100)-(3y/200)

I solve the second equation for x
x=(dy/dt)(100/3)+(y/2) and that gives me x' as
x'=(d^2y/dt^2)(100/3)+(dy/dt)(1/2)

then i plug back into the equation and collect like terms and i get
(100/3)(d^2y/dt^2)+(dy/dt)(1/2)=2-(3/100)[(dy/dt)(100/3)+(y/2)] +(y/200)

which simplifies to

(d^2y/dt^2)+(9/200)(dy/dt)+(3/10000)y=2

Then i find my two solutions for

(d^2y/dt^2)+(9/200)(dy/dt)+(3/10000)y=0

and this is were it all goes wrong...

2. Originally Posted by acosta0809
I am studying for my final and this is one of the problems in out trial final, ive gone through it about 5-6 times and i can not for the life of me get it right.

the pic and question is attached

What i have is

dx/dt=2-(3x/100)+(y/200)
dy/dt=(3x/100)-(3y/200)

I solve the second equation for x
x=(dy/dt)(100/3)+(y/2) and that gives me x' as
x'=(d^2y/dt^2)(100/3)+(dy/dt)(1/2)

then i plug back into the equation and collect like terms and i get
(100/3)(d^2y/dt^2)+(dy/dt)(1/2)=2-(3/100)[(dy/dt)(100/3)+(y/2)] +(y/200)

which simplifies to

(d^2y/dt^2)+(9/200)(dy/dt)+(3/10000)y=2

Then i find my two solutions for

(d^2y/dt^2)+(9/200)(dy/dt)+(3/10000)y=0

and this is were it all goes wrong...

Okay, why does it "all go wrong"? I haven't checked the details but your general idea is correct. Now you have a linear equation with constant coefficients with characteristic equation $r^2+ (9/200)r+ 3/10000= 0$. By the quadratic formula, $r= \frac{-(9/200)\pm\sqrt{(9/200)^2- 4(3/10000)}}{2}= 0.0225\pm 0.01436$.