Forming differential equations from words
$begingroup$
Consider two tanks, A and B, each holding 200 litres of water. A pipe pumps
water from tank A to tank B at a rate of 5 l/min. At the same time another
pipe pumps liquid from tank B to tank A at the same rate. At time $t=0$, $x_0$
kg of a chemical X is dissolved into tank A, and tank B has $y_0$ kg of the same
chemical X dissolved into it.
I got
$$begin{align*}frac{dx}{dt}&=-x_0frac{t}{40}+y_0frac{t}{40}\
frac{dy}{dt}&=-y_0frac{t}{40}+x_0frac{t}{40}end{align*}$$
Is this right? I know it seems easy but I've seen some say the signs are the other way around.
Thanks!
ordinary-differential-equations
edited Dec 6 '18 at 3:04
obscurans
1,152311
asked Dec 6 '18 at 2:09
AnoUser1AnoUser1
765
$endgroup$
add a comment |
$begingroup$
Consider two tanks, A and B, each holding 200 litres of water. A pipe pumps
water from tank A to tank B at a rate of 5 l/min. At the same time another
pipe pumps liquid from tank B to tank A at the same rate. At time $t=0$, $x_0$
kg of a chemical X is dissolved into tank A, and tank B has $y_0$ kg of the same
chemical X dissolved into it.
I got
$$begin{align*}frac{dx}{dt}&=-x_0frac{t}{40}+y_0frac{t}{40}\
frac{dy}{dt}&=-y_0frac{t}{40}+x_0frac{t}{40}end{align*}$$
Is this right? I know it seems easy but I've seen some say the signs are the other way around.
Thanks!
ordinary-differential-equations
edited Dec 6 '18 at 3:04
obscurans
1,152311
asked Dec 6 '18 at 2:09
AnoUser1AnoUser1
765
$endgroup$
add a comment |
$begingroup$
Consider two tanks, A and B, each holding 200 litres of water. A pipe pumps
water from tank A to tank B at a rate of 5 l/min. At the same time another
pipe pumps liquid from tank B to tank A at the same rate. At time $t=0$, $x_0$
kg of a chemical X is dissolved into tank A, and tank B has $y_0$ kg of the same
chemical X dissolved into it.
I got
$$begin{align*}frac{dx}{dt}&=-x_0frac{t}{40}+y_0frac{t}{40}\
frac{dy}{dt}&=-y_0frac{t}{40}+x_0frac{t}{40}end{align*}$$
Is this right? I know it seems easy but I've seen some say the signs are the other way around.
Thanks!
ordinary-differential-equations
edited Dec 6 '18 at 3:04
obscurans
1,152311
asked Dec 6 '18 at 2:09
AnoUser1AnoUser1
765
$endgroup$
Consider two tanks, A and B, each holding 200 litres of water. A pipe pumps
water from tank A to tank B at a rate of 5 l/min. At the same time another
pipe pumps liquid from tank B to tank A at the same rate. At time $t=0$, $x_0$
kg of a chemical X is dissolved into tank A, and tank B has $y_0$ kg of the same
chemical X dissolved into it.
I got
$$begin{align*}frac{dx}{dt}&=-x_0frac{t}{40}+y_0frac{t}{40}\
frac{dy}{dt}&=-y_0frac{t}{40}+x_0frac{t}{40}end{align*}$$
Is this right? I know it seems easy but I've seen some say the signs are the other way around.
Thanks!
ordinary-differential-equations
ordinary-differential-equations
edited Dec 6 '18 at 3:04
obscurans
1,152311
asked Dec 6 '18 at 2:09
AnoUser1AnoUser1
765
edited Dec 6 '18 at 3:04
obscurans
1,152311
asked Dec 6 '18 at 2:09
AnoUser1AnoUser1
765
edited Dec 6 '18 at 3:04
obscurans
1,152311
edited Dec 6 '18 at 3:04
obscurans
1,152311
edited Dec 6 '18 at 3:04
obscurans
1,152311
1,152311
asked Dec 6 '18 at 2:09
AnoUser1AnoUser1
765
asked Dec 6 '18 at 2:09
AnoUser1AnoUser1
765
asked Dec 6 '18 at 2:09
AnoUser1AnoUser1
765
765
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
For this type of problem, assume that the water in the tanks is well-mixed.
What this means is at time $t$, the concentrations of chemicals X and Y in tank A is not static, it changes as water from the other tank goes in.
Your current equation
- doesn't specify what $x$ and $y$ represent as a function of $t$
- says "the change in $x$ is some constant depending on $x_0$ and $y_0$"
- which doesn't depend on the current value of $x$ or $y$ at all
answered Dec 6 '18 at 2:44
obscuransobscurans
1,152311
$endgroup$
add a comment |
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1 Answer
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1 Answer
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votes
$begingroup$
For this type of problem, assume that the water in the tanks is well-mixed.
What this means is at time $t$, the concentrations of chemicals X and Y in tank A is not static, it changes as water from the other tank goes in.
Your current equation
- doesn't specify what $x$ and $y$ represent as a function of $t$
- says "the change in $x$ is some constant depending on $x_0$ and $y_0$"
- which doesn't depend on the current value of $x$ or $y$ at all
answered Dec 6 '18 at 2:44
obscuransobscurans
1,152311
$endgroup$
add a comment |
$begingroup$
For this type of problem, assume that the water in the tanks is well-mixed.
What this means is at time $t$, the concentrations of chemicals X and Y in tank A is not static, it changes as water from the other tank goes in.
Your current equation
- doesn't specify what $x$ and $y$ represent as a function of $t$
- says "the change in $x$ is some constant depending on $x_0$ and $y_0$"
- which doesn't depend on the current value of $x$ or $y$ at all
answered Dec 6 '18 at 2:44
obscuransobscurans
1,152311
$endgroup$
add a comment |
$begingroup$
For this type of problem, assume that the water in the tanks is well-mixed.
What this means is at time $t$, the concentrations of chemicals X and Y in tank A is not static, it changes as water from the other tank goes in.
Your current equation
- doesn't specify what $x$ and $y$ represent as a function of $t$
- says "the change in $x$ is some constant depending on $x_0$ and $y_0$"
- which doesn't depend on the current value of $x$ or $y$ at all
answered Dec 6 '18 at 2:44
obscuransobscurans
1,152311
$endgroup$
For this type of problem, assume that the water in the tanks is well-mixed.
What this means is at time $t$, the concentrations of chemicals X and Y in tank A is not static, it changes as water from the other tank goes in.
Your current equation
- doesn't specify what $x$ and $y$ represent as a function of $t$
- says "the change in $x$ is some constant depending on $x_0$ and $y_0$"
- which doesn't depend on the current value of $x$ or $y$ at all
answered Dec 6 '18 at 2:44
obscuransobscurans
1,152311
answered Dec 6 '18 at 2:44
obscuransobscurans
1,152311
answered Dec 6 '18 at 2:44
obscuransobscurans
1,152311
answered Dec 6 '18 at 2:44
obscuransobscurans
1,152311
1,152311
add a comment |
add a comment |
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