Mixing Salt and Water - First Order Differential Equations (2023)


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In this video, we use first order, linear, ordinary differential equations to solve a mixing problem. We have a 3000L tank that is being filled and drained at the same time at 10L/min.

The solution filling the tank has a salt concentration of 0.02 kg/L, while the tank has an initial quantity of salt of 15kg. Our problem is to find the amount of salt in the tank at any given time.

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Okay today, we're going to solve a mixing problem with first-order differential equations.

Now let's say that I have a tank so I have a water tank that's filled it's filled to 3000 liters let's say, the solution is draining out of the tank, straining out of the tank at 10 liters per minute.

But at the same time, there's a solution that is filling the tank also at 10 liters per minute.

All right.

So the solution level in the tank remains constant at 3,000 liters.

And that kind of makes sense because the flow rate of a solution into the tank is the same as the flow rate out of the tank.

So we have a solution than tank and it's a saltwater solution and I want to know what the amount of salt is in this tank at any given moment.

So let's start with say, 15 kilos of salt, 15 kilograms of salt in the tank.

We have a salty solution coming in.

And the concentration of the solution is coming in is 0.02 kilograms per liter.

So 20 grams per liter or 0.02 kilograms, a liter and is mixing with the salt that's already in the tank.

So let's denote the amount of salt at any given time in the tank as Y.

So Y is what we need to find.

And initially y 0, we have 15 kilos quiet time equals zero.

We've got a salty solution coming in.

So we'd expect the amount of salt in the tank to increase, but at the same time, it's flowing out as well.

So we have an unknown rate out.


So as the amount of salt that's in the tank is increasing this quantity Y is also changing over time.

Right? So we have a rate of change of the quantity.

T Y DT that is also unknown as well.

So how do we set this problem up? So let's start with dy DT dy on DT.

Okay, if we think about it, the rate of change of salt that's in the tank should be equal to the rate that's coming in - a rate that's flowing out.

And what we do know is the rate that's coming in it's equal to 0.02 kilograms, or 20 grams per liter let's convert that into a flow rate in terms of a mass over time.

So we've got 20 kilograms.

Sorry, 20 grams per liter, 0.02 kilograms per liter by 10 liters per minute.

So with the unit's here, the liters cancel out 0.02 by 10 is equal to 0.2.

So we have 0.2 kilograms per minute coming in by the same token, the raids out.

And if we think about it again, the raid out the concentration of solution, that's coming out of the tank should be the same as the concentration that's in the tank after it's been mixed properly.

Right? So we have a quantity Y coming out dissolved in 3,000 liters and it's also coming out at a rate of 10 liters per minute.

So again, cancel the unit's, one of the zeros goes from the top and bottom as well.

And we have Y over 300 kilograms per minute flowing out.

So I hope that's starting to make sense.

So the differential equation then becomes dy on DT is equal to 0.2 I'm going to leave out the unit's - Y on 300.

And if I give this expression on the right, a common denominator, we should have 300 by 0.2 minus y all over 300, which means we have 60 minus y on 300 let's call this equation 1.

An equation 1 is my linear.

First order differential equation.


So let's take a Quay, ssin 1 and let's get all of the Y's on one side onto the right hand side and all of the t's onto the left hand side.

So in other let's, separate the variables.

So we have dy on 60 minus y equals DT on 300.

And now the process is to integrate both sides.

So the right hand side integrates to log of 16 minus y and applying the reverse chain rule.

We also have to divide by the differentiation of what's in the absolute value, bars and that'll, be negative, one and that's equal to one on 300 times, t, plus the integration constant C.

Conversely, I could have used definite integrals here.

So we are integrating from y 0 to Y and we're, integrating from T equals 0 to T so it's from 0 to T, but equally I've used a indefinite integral and that's perfectly fine as well.

So we have negative of the log of 60 minus y equals T on 300, plus C.

And the next step is to isolate y so that we can find the quantity as a function of time, I'm going to exponentiate both sides of the equation.

So we have e to the negative log of 60 minus y equals e to the T on 300 plus C.

Now, one of our log laws of courses we can take whatever is in front of the logarithm and put it as a power so write that as e to the log of 60, minus y to the power of negative 1 and I can write this on the right hand side here as e to the power of T on 300 by e, to the power of C.

So that's, just using the addition law of indices, ok.

So the expression on the left hand side simplifies to these are opposite functions.

So we just left with 60 minus y to the power of negative 1 is equal to I'm just going to keep the right-hand side, the same for now so e to the power of T on 300 by e, to the power of C because now I'm going to apply a power of 1 to both sides power of negative 1 to both sides.

So 60 minus y to the power of negative 1 and I'm going to raise that again to the power of negative 1 is equal to e to the power of T on 300 by e, to the power of C to the power of negative 1 and expanding the powers in into the brackets.

We have 60 minus y is equal to e to the power of negative t on 300 by e to the power of negative C.

And if we rearrange further let's, take the 60 over to the other side, so negative y equals e to the negative T on 300 by e to the negative C, minus 60 and multiplying both sides of the equation by negative 1.

We further reduces to y equals positive, 60 minus e to the negative T on 300 now, I'm going to write e to the negative C as some constant let's call that a, and of course, it's prettier to write the a at the front.


So we're nearly done.

All that's left to do is to apply the initial condition to find our coefficient a here.

So doing that at time equals 0, we know we started with the initial amount of salt in the tank, which was 15 kilos and that's equal to the general expression, which is 60 minus a by e to the negative 0 time equals zero on 300 this expression shortens through 15 equals 60 minus a by e to the zero.

And of course, anything to the power of zero is equal to one.

So that of course means a is equal to fifteen.

Sorry, 45.


So problem solved, we have found the expression for the quantity of salt at any given time in the tank.

So Y of T is equal to 60 minus 15 sorry, minus 45 by e, to the power of T divided by 300.

Each of the pair of negative T divided by 300 all right.

So we can plug in any positive value for time.

And we should be able to precisely determine the amount of salt that's in the tank.

Let's see what this solution looks like graphically so we'll call this equation.

Two let's, see what equation two looks like graphically.


So here's a graph that I prepared earlier.

So as we can see the initial level of salt that's in the tank is 15, 15 kilos 15 kilograms as the solution starts pouring in.

We can see that there's quite an or relatively rapid increase in the solution increase in the quantity, rather of salt for the first 400 minutes, roughly the first seven hours.

But as we can see over a sufficiently long time, the amount of salt that's in the tank tapers towards 60 kilograms.

So seems will never exceed 60 kilograms of salt in a tank.

Simply by virtue that there is an outflow of the brine at the same rate that is coming in so we'll reach a steady state solution in the tank and a steady state quantity of 60 kilos.


So that's, our mixing problem done, if you have any questions feel free to ask in the comments below if I haven't made anything clear, please stay there also and I'll address that in future videos.

But hey, we have reached 200 videos.

Thank you for being part of my journey so far and watching at the moment, this channel has about six and a half thousand subscribers and nearly 1.7 million views.

And it is my goal for 2019 to double the amount of videos and double those statistics.

But more importantly is to provide good tutorials for viewers like yourself.

So if you found it useful, please like subscribe and share because I have great things planned for this channel for the future.

But for now best of luck with your studies and I'll see you on the next video.


Mixing Salt and Water - First Order Differential Equations? ›

A first order differential equation is an equation of the form F(t,y,˙y)=0. A solution of a first order differential equation is a function f(t) that makes F(t,f(t),f′(t))=0 for every value of t.

What is the rule for first order differential equations? ›

A first order differential equation is an equation of the form F(t,y,˙y)=0. A solution of a first order differential equation is a function f(t) that makes F(t,f(t),f′(t))=0 for every value of t.

What is a mixing problem system of differential equations? ›

Mixing problems are an application of separable differential equations. They're word problems that require us to create a separable differential equation based on the concentration of a substance in a tank.

What is the differential equation for water flow? ›

A general differential equation for water storage y' + CPy2 - cyk= O ib derived, lI'ith c and k constants for the given reservoir', outflow shape and type of floll', and P being the inflow hydrograph.

What is the differential equation for the amount of salt in the tank at any time? ›

The differential equation for the variation of amount of salt in a tank is given by: (dx/dt) + (x/20) = 10. Where x is in kg and t is in minutes.

Which method is used to solve the first order differential equation? ›

A homogeneous differential equation is represented by the form f(x,y)dy = g(x,y)dx only if the degree of f(x,y) and g(x,y) is the same, First order differential equations can be solved by two methods: Integrating factor and variation of constant.

What is the formula for mixing problems? ›

Solving a percent mixture problem can be done using the equation Ar = Q, where A is the amount of a solution, r is the percent concentration of a substance in the solution, and Q is the quantity of the substance in the solution.

What is an example of a mixing problem? ›

A typical mixing problem deals with the amount of salt in a mixing tank. Salt and water enter the tank at a certain rate, are mixed with what is already in the tank, and the mixture leaves at a certain rate. We want to write a differential equation to model the situation, and then solve it.

What are two types of mixture problems? ›

GMAT Quantitative: Two Types of Mixture Problems
  • Proportions in mixture problems. First, there are mixture problems that ask you to alter the proportions of a single mixture. ...
  • Combining different mixtures. The other type of mixture problem will ask you to combine two mixtures. ...
  • Solving a sample mixture problem.
Aug 24, 2019

What is Bernoulli equation for water? ›

p1+12ρv21=p2+12ρv22. Situations in which fluid flows at a constant depth are so common that this equation is often also called Bernoulli's principle, which is simply Bernoulli's equation for fluids at constant depth.

Is Bernoulli's equation a differential equation? ›

Bernoulli equations are special because they are nonlinear differential equations with known exact solutions. A notable special case of the Bernoulli equation is the logistic differential equation.

What is Darcy's law of fluid flow? ›

Darcy's law says that the discharge rate q is proportional to the gradient in hydrauolic head and the hydraulic conductivity (q = Q/A = -K*dh/dl). Definitions of aquifers, aquitards, and aquicludes and how hydraulic conductivity relates to geology.

When salt is dissolved in water what will be the change in volume and why? ›

(a) No, there will be no change in volume. The particles of common salt will occupy inter particle empty spaces present in the molecules of water.

How much salt can dissolve in water ratio? ›

At 20 °C (68 °F) one liter of water can dissolve about 357 grams of salt, a concentration of 26.3% w/w. At boiling (100 °C (212 °F)) the amount that can be dissolved in one liter of water increases to about 391 grams, a concentration of 28.1% w/w.

Does the volume of water change when salt is added? ›

This is a thought-provoking experiment with a surprising result. When sodium chloride dissolves in water to make a saturated solution there is a 2.5 per cent reduction in volume. One would never notice that in a beaker.

What are the applications of first order differential equations in real life? ›

Applications of first-order linear differential equations include determining motion of a rising or falling object with air resistance and finding current in an electrical circuit.

What is first order differential equation Euler method? ›

One of the simplest and oldest methods for approximating differential equations is known as the Euler's method. The Euler method is a first-order method, which means that the local error is proportional to the square of the step size, and the global error is proportional to the step size.

What is the difference between first order and second order differential equations? ›

The key difference between first and second order reactions is that the rate of a first order reaction depends on the first power of the reactant concentration in the rate equation whereas the rate of a second order reaction depends on the second power of the concentration term in the rate equation.

How many initial conditions are needed to solve a differential equation? ›

Again, we require two boundary conditions because of the second derivative in space, and likewise we need two initial conditions (position and slope) as a result of having a second derivative in time.

What is the solution to a differential equation without initial condition? ›

Solving the differential equation without the initial condition gives you y=sin(x)+C. Once you get the general solution, you can use the initial value to find a particular solution which satisfies the problem.

How do you calculate the mixing of solutions? ›

  1. Calculating the strength of a mixture of two different concentrations.
  2. Example 1.
  3. Method.
  4. Step 1: Calculate the amount of drug A in each concentration.
  5. Step 2: Add the two amounts together.
  6. Step 3: Add the two volumes together.
  7. Step 4: Put total amount over total volume.
  8. Student Learning Advisory Service.

What is the mixing dilution formula? ›

Solutions based on v/v measurements are liquid diluted into a liquid. Calculate appropriate v/v dilution using the formula C1V1 = C2V2 where C represents the concentration of the solute, and V represents volume in milliliters or ml.

What are the three types of mixing? ›

Mixtures can be classified on the basis of particle size into three different types: solutions, suspensions, and colloids.

What are the three types of mixing process? ›

In order to produce products in the chemical industry process, various types of mixing are required. The most common types of mixing are liquid-with-liquid mixing, solid-with-liquid mixing, liquid-with-gas mixing, and solid-with-solid mixing.

What is the mixture problem concept? ›

Mixture problems involve combining two or more things and determining some characteristic of either the ingredients or the resulting mixture. For example, we might want to know how much water to add to dilute a saline solution, or we might want to determine the percentage of concentrate in a jug of orange juice.

What is the formula for the concentration of a mixture? ›

The standard formula is C = m/V, where C is the concentration, m is the mass of the solute dissolved, and V is the total volume of the solution.

How do you solve mixture problems in Algebra 1? ›

The steps to solving mixture problems are:
  1. Use a variable (like x) for the thing that needs to be determined.
  2. Formulate an equation for the problem.
  3. Solve the problem algebraically.
  4. Write down the answer.
Nov 16, 2022

What are two examples of mixtures that are solutions? ›

Here is a brief list:
  • Salt water is formed when we mix salt (generally table salt) in water. ...
  • Sugar water is formed by mixing sugar in water.
  • Mouthwash consists of a number of chemicals dissolved in water.
  • Tincture of iodine is obtained by dissolving crystals of iodine in alcohol.

Does Bernoulli's principle apply to water? ›

Bernoulli's principle relates the pressure of a fluid to its elevation and its speed. Bernoulli's equation can be used to approximate these parameters in water, air or any fluid that has very low viscosity.

How does the magnus effect work? ›

The Magnus effect is a particular manifestation of Bernoulli's theorem: fluid pressure decreases at points where the speed of the fluid increases. In the case of a ball spinning through the air, the turning ball drags some of the air around with it.

What does Bernoulli's equation apply to? ›

The Bernoulli equation is applied to all incompressible fluid flow problems. The Bernoulli equation can be applied to devices such as the orifice meter, Venturi meter, and Pitot tube and its applications for measuring flow in open channels and inside tubes.

Can a differential equation be linear and Bernoulli? ›

When n = 0 the equation can be solved as a First Order Linear Differential Equation. When n = 1 the equation can be solved using Separation of Variables. and turning it into a linear differential equation (and then solve that).

What does Bernoulli's equation basically say? ›

Bernoulli's principle formulated by Daniel Bernoulli states that as the speed of a moving fluid increases (liquid or gas), the pressure within the fluid decreases. Although Bernoulli deduced the law, it was Leonhard Euler who derived Bernoulli's equation in its usual form in the year 1752.

What is the general solution of Bernoulli's differential equation? ›

To find the solution, change the dependent variable from y to z, where z = y1−n. This gives a differential equation in x and z that is linear, and can be solved using the integrating factor method. dy dx + P(x)y = Q(x) yn , where P and Q are functions of x, and n is a constant.

What is Darcy vs non Darcy flow? ›

Darcy Flow: Darcy Flow is the flow that satisfies the linear relationship described by Darcy's law. Non-Darcy Flow: non-Darcy Flow is the flow that has a non-linear relationship between flow rate and pressure difference and cannot be described by Darcy's law.

What is Darcy's law simple? ›

Darcy's law describes the relationship between the instantaneous rate of discharge through a porous medium and pressure drop at a distance. Using the specific sign convention, Darcy's law is expressed as: Q = -KA dh/dl. Wherein: Q is the rate of water flow.

What was Darcy's law experiment? ›

Darcy's experiments consisted of a vertical steel column, with a water inlet at one end and an outlet at the other. The water pressure was controlled at the inlet and outlet ends of the column using reservoirs with constant water levels (Figure 25) (denoted h1 and h2).

What change happens when salt dissolves in water? ›

Dissolving salt in water is a chemical change.

What will happen when we add salt more than the quantity of water? ›

If you add salt to water, you raise the water's boiling point, or the temperature at which it will boil. The temperature needed to boil will increase about 0.5 C for every 58 grams of dissolved salt per kilogram of water.

What will happen if you put a handful of salt into a glass full of water will the water overflow Why? ›

why? Sodium chloride is a hygroscopic salt that means it absorbs water, so that the level of water goes down when we put a handful of sodium chloride in a glass of water.

What is salt mixed with water called? ›

Saltwater acts as if it were a single substance even though it contains two substances—salt and water. Saltwater is a homogeneous mixture, or a solution.

How much salt will dissolve in 1 gallon of water? ›

One gallon of water will dissolve 3 pounds of salt. To dissolve 8 pounds of salt, at least 3 gallons of water should be in the brine tank.

What is the rule of solubility of salt in water? ›

Salts containing nitrate ion (NO3-) are generally soluble. Salts containing Cl -, Br -, or I - are generally soluble. Important exceptions to this rule are halide salts of Ag+, Pb2+, and (Hg2)2+.

What is the effect of mixing salt and water? ›

When salt is mixed with water, the salt dissolves because the covalent bonds of water are stronger than the ionic bonds in the salt molecules.

Does adding salt to water increase or decrease density? ›

When salt is dissolved in fresh water, the density of the water increases because the mass of the water increases.

Does dissolving increase volume? ›

When sugar dissolves in water, the sugar molecules take up space between the water molecules. Thus, they do not occupy any extra space. Hence volume of solution does not change. Q.

What is the first order condition? ›

The first-order condition recommends that we add workers to the production process up to the point where the last worker's marginal product is equal to his wage (or cost). In addition, a second characteristic of a maximum is that the second derivative is negative (or nonpositive).

How do you know if a differential equation is first or second order? ›

If a1(x) and a2(x) are constant, then (2) has constant coefficients. is second order, linear, non homogeneous and with constant coefficients. y + x2y = ex is first order, linear, non homogeneous. yy + y = 0 is non linear, second order, homogeneous.

What is the initial condition of a differential equation? ›

An initial condition is an extra bit of information about a differential equation that tells you the value of the function at a particular point. Differential equations with initial conditions are commonly called initial value problems.

What is an example of a first order derivative? ›

Derivative – First Order. The first order derivative of a function represents the rate of change of one variable with respect to another variable. For example, in Physics we define the velocity of a body as the rate of change of the location of the body with respect to time.

What is the difference between first order condition and second order condition? ›

Another difference with the first-order condition is that the second-order condition distinguishes minima from maxima: at a local maximum, the Hessian must be negative semidefinite, while the first-order condition applies to any extremum (a minimum or a maximum).

What are the first order necessary conditions for optimality? ›

the first-order optimality measure is the infinity norm (meaning maximum absolute value) of ∇f(x), which is: first-order optimality measure = max i | ( ∇ f ( x ) ) i | = ‖ ∇ f ( x ) ‖ ∞ . This measure of optimality is based on the familiar condition for a smooth function to achieve a minimum: its gradient must be zero.

How do you distinguish between characteristics of first and second-order differential equations? ›

The order of the highest order derivative present in the differential equation is called the order of the equation. If the order of the differential equation is 1, then it is called the first order. If the order of the equation is 2, then it is called a second-order, and so on.

What is the difference between the first and second-order derivatives? ›

Graphically the first derivative represents the slope of the function at a point, and the second derivative describes how the slope changes over the independent variable in the graph. For a function having a variable slope, the second derivative explains the curvature of the given graph.

What is the difference between first order and second-order de? ›

As for a first-order difference equation, we can find a solution of a second-order difference equation by successive calculation. The only difference is that for a second-order equation we need the values of x for two values of t, rather than one, to get the process started.

Does the initial conditions of a differential equation affect the solution? ›

In both differential equations in continuous time and difference equations in discrete time, initial conditions affect the value of the dynamic variables (state variables) at any future time.

What is the condition of solution in differential equation? ›

A differential equation is an equation involving an unknown function y=f(x) and one or more of its derivatives. A solution to a differential equation is a function y=f(x) that satisfies the differential equation when f and its derivatives are substituted into the equation.

What makes a first order differential equation separable? ›

Definition. A first order differential equation is separable if it can be written in one of the following forms: dydx=f(x,y)=g(x)h(y),dydx=f(x,y)=h(y)g(x).

How do you tell if a differential equation is linear or nonlinear? ›

In a differential equation, when the variables and their derivatives are only multiplied by constants, then the equation is linear. The variables and their derivatives must always appear as a simple first power.

What makes a first order differential equation homogeneous? ›

A first order differential equation is homogeneous if it takes the form: dydx=F(yx), where F(yx) F ( y x ) is a homogeneous function. In this context homogeneous is used to mean a function of x and y that is left unchanged by multiplying both arguments by a constant, i.e.

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