3.4: Separable Equations - Mathematics

3.4: Separable Equations - Mathematics

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Learning Objectives

  • Use separation of variables to solve a differential equation.
  • Solve applications using separation of variables.

We now examine a solution technique for finding exact solutions to a class of differential equations known as separable differential equations. These equations are common in a wide variety of disciplines, including physics, chemistry, and engineering. We illustrate a few applications at the end of the section.

Separation of Variables

We start with a definition and some examples.

Definition: Separable Differential Equations

A separable differential equation is any equation that can be written in the form

[y'=f(x)g(y). label{sep}]

The term ‘separable’ refers to the fact that the right-hand side of Equation ef{sep} can be separated into a function of (x) times a function of (y). Examples of separable differential equations include

[ egin{align} y' =(x^2−4)(3y+2) label{eq1} [4pt] y' =6x^2+4x label{eq2}[4pt] y' =sec y+ an y label{eq3} [4pt] y' =xy+3x−2y−6. label{eq4} end{align}]

Equation ef{eq2} is separable with (f(x)=6x^2+4x) and (g(y)=1), Equation ef{eq3} is separable with (f(x)=1) and (g(y)=sec y+ an y,) and the right-hand side of Equation ef{eq4} can be factored as ((x+3)(y−2)), so it is separable as well. Equation ef{eq3} is also called an autonomous differential equation because the right-hand side of the equation is a function of (y) alone. If a differential equation is separable, then it is possible to solve the equation using the method of separation of variables.

Problem-Solving Strategy: Separation of Variables

  1. Check for any values of (y) that make (g(y)=0.) These correspond to constant solutions.
  2. Rewrite the differential equation in the form [ dfrac{dy}{g(y)}=f(x)dx.]
  3. Integrate both sides of the equation.
  4. Solve the resulting equation for (y) if possible.
  5. If an initial condition exists, substitute the appropriate values for (x) and (y) into the equation and solve for the constant.

Note that Step 4 states “Solve the resulting equation for (y) if possible.” It is not always possible to obtain (y) as an explicit function of (x). Quite often we have to be satisfied with finding y as an implicit function of (x).

Example (PageIndex{1}): Using Separation of Variables

Find a general solution to the differential equation (y'=(x^2−4)(3y+2)) using the method of separation of variables.


Follow the five-step method of separation of variables.

1. In this example, (f(x)=x^2−4) and (g(y)=3y+2). Setting (g(y)=0) gives (y=−dfrac{2}{3}) as a constant solution.

2. Rewrite the differential equation in the form

[ dfrac{dy}{3y+2}=(x^2−4)dx.]

3. Integrate both sides of the equation:

[ ∫dfrac{dy}{3y+2}=∫(x^2−4)dx.]

Let (u=3y+2). Then (du=3dfrac{dy}{dx}dx), so the equation becomes

[ dfrac{1}{3}∫dfrac{1}{u}du=dfrac{1}{3}x^3−4x+C]

[ dfrac{1}{3}ln|u|=dfrac{1}{3}x^3−4x+C]

[ dfrac{1}{3}ln|3y+2|=dfrac{1}{3}x^3−4x+C.]

4. To solve this equation for (y), first multiply both sides of the equation by (3).

[ ln|3y+2|=x^3−12x+3C]

Now we use some logic in dealing with the constant (C). Since (C) represents an arbitrary constant, (3C) also represents an arbitrary constant. If we call the second arbitrary constant (C_1), the equation becomes

[ ln|3y+2|=x^3−12x+C_1.]

Now exponentiate both sides of the equation (i.e., make each side of the equation the exponent for the base (e)).

[ egin{align} e^{ln|3y+2|} =e^{x^3−12x+C_1} |3y+2| =e^{C_1}e^{x^3−12x} end{align}]

Again define a new constant (C_2=e^{c_1}) (note that (C_2>0)):

[ |3y+2|=C_2e^{x^3−12x}.]

This corresponds to two separate equations:




The solution to either equation can be written in the form


Since (C_2>0), it does not matter whether we use plus or minus, so the constant can actually have either sign. Furthermore, the subscript on the constant (C) is entirely arbitrary, and can be dropped. Therefore the solution can be written as

[ y=dfrac{−2+Ce^{x^3−12x}}{3}.]

5. No initial condition is imposed, so we are finished.

Exercise (PageIndex{1})

Use the method of separation of variables to find a general solution to the differential equation

[ y'=2xy+3y−4x−6. onumber]


First factor the right-hand side of the equation by grouping, then use the five-step strategy of separation of variables.


[ y=2+Ce^{x^2+3x} onumber]

Example (PageIndex{2}): Solving an Initial-Value Problem

Using the method of separation of variables, solve the initial-value problem

[ y'=(2x+3)(y^2−4),y(0)=−1.]


Follow the five-step method of separation of variables.

1. In this example, (f(x)=2x+3) and (g(y)=y^2−4). Setting (g(y)=0) gives (y=±2) as constant solutions.

2. Divide both sides of the equation by (y^2−4) and multiply by (dx). This gives the equation


3. Next integrate both sides:

[∫dfrac{1}{y^2−4}dy=∫(2x+3)dx. label{Ex2.2}]

To evaluate the left-hand side, use the method of partial fraction decomposition. This leads to the identity

[dfrac{1}{y^2−4}=dfrac{1}{4}left(dfrac{1}{y−2}−dfrac{1}{y+2} ight).]

Then Equation ef{Ex2.2} becomes

[dfrac{1}{4}∫left(dfrac{1}{y−2}−dfrac{1}{y+2} ight)dy=∫(2x+3)dx]

[dfrac{1}{4}left (ln|y−2|−ln|y+2| ight)=x^2+3x+C.]

Multiplying both sides of this equation by (4) and replacing (4C) with (C_1) gives


[ln left|dfrac{y−2}{y+2} ight|=4x^2+12x+C_1.]

4. It is possible to solve this equation for y. First exponentiate both sides of the equation and define (C_2=e^{C_1}):

[left|dfrac{y−2}{y+2} ight|=C_2e^{4x^2+12x}.]

Next we can remove the absolute value and let (C_2) be either positive or negative. Then multiply both sides by (y+2).



Now collect all terms involving y on one side of the equation, and solve for y:




5. To determine the value of (C_2), substitute (x=0) and (y=−1) into the general solution. Alternatively, we can put the same values into an earlier equation, namely the equation (dfrac{y−2}{y+2}=C_2e^{4x^2+12}). This is much easier to solve for (C_2):




Therefore the solution to the initial-value problem is


A graph of this solution appears in Figure (PageIndex{1}).

Exercise (PageIndex{2})

Find the solution to the initial-value problem

[ 6y'=(2x+1)(y^2−2y−8) onumber]

with (y(0)=−3) using the method of separation of variables.


Follow the steps for separation of variables to solve the initial-value problem.


[ y=dfrac{4+14e^{x^2+x}}{1−7e^{x^2+x}} onumber]

Applications of Separation of Variables

Many interesting problems can be described by separable equations. We illustrate two types of problems: solution concentrations and Newton’s law of cooling.

Solution concentrations

Consider a tank being filled with a salt solution. We would like to determine the amount of salt present in the tank as a function of time. We can apply the process of separation of variables to solve this problem and similar problems involving solution concentrations.

Example (PageIndex{3}): Determining Salt Concentration over Time

A tank containing (100,L) of a brine solution initially has (4,kg) of salt dissolved in the solution. At time (t=0), another brine solution flows into the tank at a rate of (2,L/min). This brine solution contains a concentration of (0.5,kg/L) of salt. At the same time, a stopcock is opened at the bottom of the tank, allowing the combined solution to flow out at a rate of (2,L/min), so that the level of liquid in the tank remains constant (Figure (PageIndex{2})). Find the amount of salt in the tank as a function of time (measured in minutes), and find the limiting amount of salt in the tank, assuming that the solution in the tank is well mixed at all times.


First we define a function (u(t)) that represents the amount of salt in kilograms in the tank as a function of time. Then (dfrac{du}{dt}) represents the rate at which the amount of salt in the tank changes as a function of time. Also, (u(0)) represents the amount of salt in the tank at time (t=0), which is (4) kilograms.

The general setup for the differential equation we will solve is of the form


INFLOW RATE represents the rate at which salt enters the tank, and OUTFLOW RATE represents the rate at which salt leaves the tank. Because solution enters the tank at a rate of (2)L/min, and each liter of solution contains (0.5) kilogram of salt, every minute (2(0.5)=1kilogram) of salt enters the tank. Therefore INFLOW RATE = (1).

To calculate the rate at which salt leaves the tank, we need the concentration of salt in the tank at any point in time. Since the actual amount of salt varies over time, so does the concentration of salt. However, the volume of the solution remains fixed at 100 liters. The number of kilograms of salt in the tank at time (t) is equal to (u(t)). Thus, the concentration of salt is (dfrac{u(t)}{100} kg/L), and the solution leaves the tank at a rate of (2) L/min. Therefore salt leaves the tank at a rate of (dfrac{u(t)}{100}⋅2=dfrac{u(t)}{50}) kg/min, and OUTFLOW RATE is equal to (dfrac{u(t)}{50}). Therefore the differential equation becomes (dfrac{du}{dt}=1−dfrac{u}{50}), and the initial condition is (u(0)=4.) The initial-value problem to be solved is


The differential equation is a separable equation, so we can apply the five-step strategy for solution.

Step 1. Setting (1−dfrac{u}{50}=0) gives (u=50) as a constant solution. Since the initial amount of salt in the tank is (4) kilograms, this solution does not apply.

Step 2. Rewrite the equation as


Then multiply both sides by (dt) and divide both sides by (50−u:)


Step 3. Integrate both sides:

[egin{align} ∫dfrac{du}{50−u} =∫dfrac{dt}{50} −ln|50−u| =dfrac{t}{50}+C. end{align}]

Step 4. Solve for (u(t)):




Eliminate the absolute value by allowing the constant to be either positive or negative:


Finally, solve for (u(t)):


Step 5. Solve for (C_1):

[egin{align} u(0) =50−C_1e^{−0/50} 4 =50−C_1 C_1 =46. end{align}]

The solution to the initial value problem is (u(t)=50−46e^{−t/50}.) To find the limiting amount of salt in the tank, take the limit as (t) approaches infinity:

[egin{align} lim_{t→∞}u(t) =50−46e^{−t/50} =50−46(0)=50. end{align}]

Note that this was the constant solution to the differential equation. If the initial amount of salt in the tank is (50) kilograms, then it remains constant. If it starts at less than (50) kilograms, then it approaches 50 kilograms over time.

Exercise (PageIndex{3})

A tank contains (3) kilograms of salt dissolved in (75) liters of water. A salt solution of (0.4,kg salt/L) is pumped into the tank at a rate of (6,L/min) and is drained at the same rate. Solve for the salt concentration at time (t). Assume the tank is well mixed at all times.


Follow the steps in Example (PageIndex{3}) and determine an expression for INFLOW and OUTFLOW. Formulate an initial-value problem, and then solve it.

Initial value problem:

[ dfrac{du}{dt}=2.4−dfrac{2u}{25},, u(0)=3 onumber]


[u(t)=30−27e^{−t/50} onumber]

Newton’s law of Cooling

Newton’s law of cooling states that the rate of change of an object’s temperature is proportional to the difference between its own temperature and the ambient temperature (i.e., the temperature of its surroundings). If we let (T(t)) represent the temperature of an object as a function of time, then (dfrac{dT}{dt}) represents the rate at which that temperature changes. The temperature of the object’s surroundings can be represented by (T_s). Then Newton’s law of cooling can be written in the form

[ dfrac{dT}{dt}=k(T(t)−T_s)]

or simply

[ dfrac{dT}{dt}=k(T−T_s).]

The temperature of the object at the beginning of any experiment is the initial value for the initial-value problem. We call this temperature (T_0). Therefore the initial-value problem that needs to be solved takes the form

[ dfrac{dT}{dt}=k(T−T_s) label{newton}]

with (T(0)=T_0), where (k) is a constant that needs to be either given or determined in the context of the problem. We use these equations in Example (PageIndex{4}).

Example (PageIndex{4}): Waiting for a Pizza to Cool

A pizza is removed from the oven after baking thoroughly, and the temperature of the oven is (350°F.) The temperature of the kitchen is (75°F), and after (5) minutes the temperature of the pizza is (340°F). We would like to wait until the temperature of the pizza reaches (300°F) before cutting and serving it (Figure (PageIndex{3})). How much longer will we have to wait?


The ambient temperature (surrounding temperature) is (75°F), so (T_s=75). The temperature of the pizza when it comes out of the oven is (350°F), which is the initial temperature (i.e., initial value), so (T_0=350). Therefore Equation ef{newton} becomes

[dfrac{dT}{dt}=k(T−75) ]

with (T(0)=350.)

To solve the differential equation, we use the five-step technique for solving separable equations.

1. Setting the right-hand side equal to zero gives (T=75) as a constant solution. Since the pizza starts at (350°F,) this is not the solution we are seeking.

2. Rewrite the differential equation by multiplying both sides by (dt) and dividing both sides by (T−75):

[dfrac{dT}{T−75}=kdt. onumber]

3. Integrate both sides:

[egin{align} ∫dfrac{dT}{T−75} =∫kdt onumber ln|T−75| =kt+C. onumber end{align} ]

4. Solve for (T) by first exponentiating both sides:

[egin{align}e^{ln|T−75|} =e^{kt+C} onumber |T−75| =C_1e^{kt} onumber T−75 =C_1e^{kt} onumber T(t) =75+C_1e^{kt}. onumber end{align} ]

5. Solve for (C_1) by using the initial condition (T(0)=350:)

[egin{align}T(t) =75+C_1e^{kt} onumber T(0) =75+C_1e^{k(0)} onumber 350 =75+C_1 onumber C_1 =275. onumber end{align} ]

Therefore the solution to the initial-value problem is

[T(t)=75+275e^{kt}. onumber]

To determine the value of (k), we need to use the fact that after (5) minutes the temperature of the pizza is (340°F). Therefore (T(5)=340.) Substituting this information into the solution to the initial-value problem, we have

[T(t)=75+275e^{kt} onumber]

[T(5)=340=75+275e^{5k} onumber]

[265=275e^{5k} onumber]

[e^{5k}=dfrac{53}{55} onumber]

[ln e^{5k}=ln(dfrac{53}{55}) onumber]

[5k=ln(dfrac{53}{55}) onumber]

[k=dfrac{1}{5}ln(dfrac{53}{55})≈−0.007408. onumber]

So now we have (T(t)=75+275e^{−0.007048t}.) When is the temperature (300°F)? Solving for t, we find

[T(t)=75+275e^{−0.007048t} onumber]

[300=75+275e^{−0.007048t} onumber]

[225=275e^{−0.007048t} onumber]

[e^{−0.007048t}=dfrac{9}{11} onumber]

[ln e^{−0.007048t}=lndfrac{9}{11} onumber]

[−0.007048t=lndfrac{9}{11} onumber]

[t=−dfrac{1}{0.007048}lndfrac{9}{11}≈28.5. onumber]

Therefore we need to wait an additional (23.5) minutes (after the temperature of the pizza reached (340°F)). That should be just enough time to finish this calculation.

Exercise (PageIndex{4})

A cake is removed from the oven after baking thoroughly, and the temperature of the oven is (450°F). The temperature of the kitchen is (70°F), and after (10) minutes the temperature of the cake is (430°F).

  1. Write the appropriate initial-value problem to describe this situation.
  2. Solve the initial-value problem for (T(t)).
  3. How long will it take until the temperature of the cake is within (5°F) of room temperature?

Determine the values of (T_s) and (T_0) then use Equation ef{newton}.

Answer a

Initial-value problem [dfrac{dT}{dt}=k(T−70),T(0)=450 onumber]

Answer b

[T(t)=70+380e^{kt} onumber]

Answer c

Approximately (114) minutes.

Key Concepts

  • A separable differential equation is any equation that can be written in the form (y'=f(x)g(y).)
  • The method of separation of variables is used to find the general solution to a separable differential equation.

Key Equations

  • Separable differential equation


  • Solution concentration


  • Newton’s law of cooling



autonomous differential equation
an equation in which the right-hand side is a function of (y) alone
separable differential equation
any equation that can be written in the form (y'=f(x)g(y))
separation of variables
a method used to solve a separable differential equation

Separable Space

A topological space $left( ight)$ is said to be a separable space if it has a countable dense subset in $X$ i.e., $A subseteq X$, $overline A = X$, or $A cup U e phi $, where $U$ is an open set.

In other words, a space $X$ is said to be a separable space if there is a subset $A$ of $X$ such that (1) $A$ is countable (2) $overline A = X$ ($A$ is dense in$X$).

Let $X = left < <1,2,3,4,5> ight>$ be a non-empty set and $ au = left< ,left < <3,4> ight>,left < <2,3> ight>,left < <2,3,4> ight>> ight>$ be a topology defined on $X$. Suppose a subset $A = left < <1,3,5> ight> subseteq X$. The closed sets are $X,phi ,left < <1,2,4,5> ight>,left < <1,2,5> ight>,left < <1,4,5> ight>,left < <1,5> ight>$. Now we have $overline A = X$. Since $A$ is finite and dense in $X$, then $X$ is a separable space.

Consider that the set of rational numbers $mathbb$ a subset of $mathbb$ (with usual topology), then the only closed set containing $mathbb$ is $mathbb$, which shows that $overline = mathbb$. Since $mathbb$ is dense in $mathbb$, then $mathbb$ is also separable in $mathbb$. However, the set of irrational numbers is dense in $mathbb$ but not countable.

• Every second countable space is a separable space.
• Every separable space is not a second countable space.
• Every separable metric space is a second countable space.
• The continuous image of a separable space is separable.

Many problems involving separable differential equations are word problems. These problems require the additional step of translating a statement into a differential equation.

When reading a sentence that relates a function to one of its derivatives, it's important to extract the correct meaning to give rise to a differential equation. The key is to search for phrases like "rate of change" because they indicate that there is a derivative involved. In fact, the term "derivative" and phrase "rate of change" are synonymous, so these should be kept in mind when constructing a differential equation that models a word problem.

  • The rate of growth of a bacteria population is directly proportional to the current bacteria population.

  • The rate of change of an object's temperature is directly proportional to the difference of the object's current temperature and the temperature of the surrounding environment.

  • The force felt by an object in free-fall is the difference of its weight and drag force, where the drag force is proportional to the object's current velocity.

Once a word problem has been written as a differential equation, it can be solved using the techniques of the previous section.

The exact same approach can be used on other problems as well.

The rate of growth of a beanstalk is proportional to the square root of its current height. If the height is 100 feet initially and it grows to 400 feet after 5 days, how tall will it be after 20 more days?

Let the height of the beanstalk (in feet) be denoted by h h h , and the time (in days) by t t t . Then The rate of growth of a beanstalk ⏟ d h d t is ⏟ = proportional to ⏟ k the square root of its current height. ⏟ h underbrace < ext> _< huge< frac< dh > < dt >> > underbrace < ext< is >> _ < huge< = >> underbrace < ext< proportional to >> _ < huge< k >> underbrace < ext < the square root of its current height.>> _ > > d t d h ​

The rate of growth of a beanstalk ​ ​ =

is ​ ​ k

proportional to ​ ​ h

the square root of its current height. ​ ​

Now, d h h = k d t ⟹ ∫ d h h = ∫ k d t ⟹ 2 h = k t + C . frac< dh > < sqrt< h >> = k dt implies int frac< dh > < sqrt< h >> = int k dt implies 2 sqrt < h >= kt + C. h

​ d h ​ = k d t ⟹ ∫ h

​ d h ​ = ∫ k d t ⟹ 2 h

​ = k t + C .

Since h = 100 h = 100 h = 1 0 0 at t = 0 t = 0 t = 0 and h = 400 h = 400 h = 4 0 0 at t = 5 t = 5 t = 5 , we have

< 2 100 = k × 0 + C 2 400 = k × 5 + C . egin 2 sqrt < 100 >& = k imes 0 + C 2 sqrt < 400 >& = k imes 5 + C. end < 2 1 0 0

​ 2 4 0 0

​ ​ = k × 0 + C = k × 5 + C . ​

From the first equation we have C = 20 C = 20 C = 2 0 . Substituting this into the second equation gives 40 = 5 k + 20 40 = 5k + 20 4 0 = 5 k + 2 0 , which implies k = 4 k = 4 k = 4 .

Thus, we now have the particular solution for the differential equation: 2 h = 4 t + 20 ⟹ h = ( 2 t + 10 ) 2 . 2 sqrt < h >= 4t + 20 implies h = (2t + 10) ^ < 2 >. 2 h

​ = 4 t + 2 0 ⟹ h = ( 2 t + 1 0 ) 2 .

Now finding the height of the beanstalk after 20 more days is very easy. We only need to find the height of the beanstalk at t = 5 + 20 = 25 days t = 5 + 20 = 25 ext < days>t = 5 + 2 0 = 2 5 days . Simply substitute t = 25 t = 25 t = 2 5 in the equation and obtain: h = ( 2 × 25 + 10 ) 2 = 6 0 2 = 3600 (feet) . h = ( 2 imes 25 + 10 ) ^ < 2 >= 60 ^ < 2 >= 3600 ext < (feet)>. h = ( 2 × 2 5 + 1 0 ) 2 = 6 0 2 = 3 6 0 0 (feet) .

The population of a country grows at a rate proportional to the size of the population. If the population doubles in 50 years, in how many years (from now) will it triple?


Suppose a differential equation can be written in the form

which we can write more simply by letting y = f ( x ) :

As long as h(y) ≠ 0, we can rearrange terms to obtain:

so that the two variables x and y have been separated. dx (and dy) can be viewed, at a simple level, as just a convenient notation, which provides a handy mnemonic aid for assisting with manipulations. A formal definition of dx as a differential (infinitesimal) is somewhat advanced.

Alternative notation Edit

Those who dislike Leibniz's notation may prefer to write this as

but that fails to make it quite as obvious why this is called "separation of variables". Integrating both sides of the equation with respect to x , we have

If one can evaluate the two integrals, one can find a solution to the differential equation. Observe that this process effectively allows us to treat the derivative d y d x >> as a fraction which can be separated. This allows us to solve separable differential equations more conveniently, as demonstrated in the example below.

(Note that we do not need to use two constants of integration, in equation (A1) as in

∫ 1 h ( y ) d y + C 1 = ∫ g ( x ) d x + C 2 , >,dy+C_<1>=int g(x),dx+C_<2>,>

Example Edit

Population growth is often modeled by the differential equation

Separation of variables may be used to solve this differential equation.

To evaluate the integral on the left side, we simplify the fraction

and then, we decompose the fraction into partial fractions

Therefore, the solution to the logistic equation is

Generalization of separable ODEs to the nth order Edit

Much like one can speak of a separable first-order ODE, one can speak of a separable second-order, third-order or nth-order ODE. Consider the separable first-order ODE:

The derivative can alternatively be written the following way to underscore that it is an operator working on the unknown function, y:

Thus, when one separates variables for first-order equations, one in fact moves the dx denominator of the operator to the side with the x variable, and the d(y) is left on the side with the y variable. The second-derivative operator, by analogy, breaks down as follows:

The third-, fourth- and nth-derivative operators break down in the same way. Thus, much like a first-order separable ODE is reducible to the form

a separable second-order ODE is reducible to the form

and an nth-order separable ODE is reducible to

Example Edit

Consider the simple nonlinear second-order differential equation:

The method of separation of variables is also used to solve a wide range of linear partial differential equations with boundary and initial conditions, such as the heat equation, wave equation, Laplace equation, Helmholtz equation and biharmonic equation.

The analytical method of separation of variables for solving partial differential equations has also been generalized into a computational method of decomposition in invariant structures that can be used to solve systems of partial differential equations. [1]

Example: homogeneous case Edit

Consider the one-dimensional heat equation. The equation is

The variable u denotes temperature. The boundary condition is homogeneous, that is

Let us attempt to find a solution which is not identically zero satisfying the boundary conditions but with the following property: u is a product in which the dependence of u on x, t is separated, that is:

Substituting u back into equation (1) and using the product rule,

Since the right hand side depends only on x and the left hand side only on t, both sides are equal to some constant value −λ. Thus:

λ here is the eigenvalue for both differential operators, and T(t) and X(x) are corresponding eigenfunctions.

We will now show that solutions for X(x) for values of λ ≤ 0 cannot occur:

Suppose that λ < 0. Then there exist real numbers B, C such that

and therefore B = 0 = C which implies u is identically 0.

Suppose that λ = 0. Then there exist real numbers B, C such that

From (7) we conclude in the same manner as in 1 that u is identically 0.

Therefore, it must be the case that λ > 0. Then there exist real numbers A, B, C such that

From (7) we get C = 0 and that for some positive integer n,

This solves the heat equation in the special case that the dependence of u has the special form of (3).

In general, the sum of solutions to (1) which satisfy the boundary conditions (2) also satisfies (1) and (3). Hence a complete solution can be given as

where Dn are coefficients determined by initial condition.

Given the initial condition

This is the sine series expansion of f(x). Multiplying both sides with sin ⁡ n π x L < extstyle sin >> and integrating over [0, L] result in

Example: nonhomogeneous case Edit

Suppose the equation is nonhomogeneous,

with the boundary condition the same as (2).

Expand h(x,t), u(x,t) and f(x) into

where hn(t) and bn can be calculated by integration, while un(t) is to be determined.

Substitute (9) and (10) back to (8) and considering the orthogonality of sine functions we get

which are a sequence of linear differential equations that can be readily solved with, for instance, Laplace transform, or Integrating factor. Finally, we can get

If the boundary condition is nonhomogeneous, then the expansion of (9) and (10) is no longer valid. One has to find a function v that satisfies the boundary condition only, and subtract it from u. The function u-v then satisfies homogeneous boundary condition, and can be solved with the above method.

Example: mixed derivatives Edit

For some equations involving mixed derivatives, the equation does not separate as easily as the heat equation did in the first example above, but nonetheless separation of variables may still be applied. Consider the two-dimensional biharmonic equation

Proceeding in the usual manner, we look for solutions of the form

and we obtain the equation

Writing this equation in the form

E ( x ) + F ( x ) G ( y ) + H ( y ) = 0 ,

we see that the derivative with respect to x and y eliminates the first and last terms, so that

Curvilinear coordinates Edit

In orthogonal curvilinear coordinates, separation of variables can still be used, but in some details different from that in Cartesian coordinates. For instance, regularity or periodic condition may determine the eigenvalues in place of boundary conditions. See spherical harmonics for example.

Partial differential equations

For many PDEs, such as the wave equation, Helmholtz equation and Schrodinger equation, the applicability of separation of variables is a result of the spectral theorem. In some cases, separation of variables may not be possible. Separation of variables may be possible in some coordinate systems but not others, [2] and which coordinate systems allow for separation depends on the symmetry properties of the equation. [3] Below is an outline of an argument demonstrating the applicability of the method to certain linear equations, although the precise method may differ in individual cases (for instance in the biharmonic equation above).

Hence, the spectral theorem ensures that the separation of variables will (when it is possible) find all the solutions.

The matrix form of the separation of variables is the Kronecker sum.

As an example we consider the 2D discrete Laplacian on a regular grid:

Some mathematical programs are able to do separation of variables: Xcas [5] among others.


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3.4: Separable Equations - Mathematics

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Mathematical Expression Editor

We solve separable differential equations and initial value problems.

Differential Equations

A differential equation is an equation that involves one or more derivatives of an unknown function. Solving a differential equation entails determining the unknown function.

An initial value problem is a differential equation along with other information about the solution, usually the value of the function at a point. The purpose of the initial value is to determine one specific solution of the differential equation, in the event that there was more than one solution.

The general solution of the differential equation is

The solution to the initial value problem is

Separable Differential Equations

A separable differential equation is a differential equation that can be put in the form . To solve such an equation, we separate the variables by moving the ’s to one side and the ’s to the other, then integrate both sides with respect to and solve for . In general, the process goes as follows: Let for convenience and we have

Integrating both sides with respect to using a the substitution , we have and so Now we solve for algebraically to get the final answer. To simplify the computations, we will use instead of and then make a slight abuse of notation to get to the same end result. From the beginning: again using for convenience. Now we write this last equation in differential form and put an integral sign on both sides(!): yielding as before.

Table of Contents

This book presents a comprehensive introduction to the theory of separable algebras over commutative rings. After a thorough introduction to the general theory, the fundamental roles played by separable algebras are explored. For example, Azumaya algebras, the henselization of local rings, and Galois theory are rigorously introduced and treated. Interwoven throughout these applications is the important notion of étale algebras. Essential connections are drawn between the theory of separable algebras and Morita theory, the theory of faithfully flat descent, cohomology, derivations, differentials, reflexive lattices, maximal orders, and class groups.

The text is accessible to graduate students who have finished a first course in algebra, and it includes necessary foundational material, useful exercises, and many nontrivial examples.

Graduate students and researchers interested in algebra.

The book is neatly arranged. It can be used as a textbook for self-study and as a reference text for most of the topics related to separability. It will be a valuable resource for students and researchers and has the potential to be a standard reference on separable algebras for many years.

-- Wolfgang Rump, Mathematical Reviews

The thorough and comprehensive treatment of separable, Azumaya, and tale algebras, Hensel rings, the Galois theory of rings, and Galois cohomology of rings makes the book under review an indispensable reference for the graduate student interested in these topics. As an added bonus, the book comes with a rich, 155 item, bibliography, well-chosen examples, calculations, and sets of exercises in each chapter, which makes this book an excellent textbook for self-study or for a topics course on separable algebras.

-- Felipe Zaldivar, MAA Reviews

Numerical solution of non-separable elliptic equations by the iterative application of FFT methods

A method for the numerical solution of non-separable (self-adjoint) elliptic equations is described in which the basic approach is the iterative application of direct methods. Such equations may be transformed into Helmholtz form and this Helmholtz problem is solved by the iterative application of FFT methods. An equation which is ‘near’ (in some sense) to the Helmholtz equation is appropriately chosen from the general class of equations soluble directly by FFT methods (see, for example, Le Bail, 1972) and the iteration (of block-Jacobi form) consists of corrections to the relevant Fourier harmonic amplitudes of the solution of this ‘nearby’ equation. It is also shown that the method is equivalent to a D' Yakonov-Gunn iteration [D' Yakonov (1961), Gunn(1964)] with a particular choice of iteration parameter and it is well known that, for self-adjoint problems with smooth coefficients, this form of iteration has a convergence rate which is essentially independent of grid-size.

In the Concus and Golub (1973) method for solving non-separable elliptic equations such equations are transformed to Helmholtz form and Poisson's equation is employed as the ‘nearby’ equation. At each iterative stage this equation is solved by the Bunemann (cyclic reduction) algorithm. Their method also uses shifted iterations to improve convergence rates and we adopt a similar approach in the present study. However, our choice of nearby equation allows the use of a form of variable shift (in addition to the constant shift used by Concus et al.,) and it will be demonstrated that the use of such a variable shift can considerably improve rates of convergence in some examples.

Numerical results are presented for illustrative examples in the unit square with Dirichlet boundary conditions and the general computational behaviour of the method was found to be in very good agreement with that predicted by a theoretical analysis. The general iterative approach may be extended to more general linear elliptic equations.

1 Answer 1

Separation of variables works on regions that are rectangular in some particular coordinate system. The underlying operator must be separable in that coordinate system. For the Laplacian, that generally means an orthogonal coordinate system such as spherical coordinates, cylindrical coordinates, elliptic coordinates, etc.. A rectangular region in spherical coordinates can be a sphere, a spherical thick shell, a wedge, etc.. There are a couple of dozen such coordinate systems. You can add various potentials, provided their dependence is also separable (usually meaning that the it depends on only one of the coordinate variables.)

So the number of configurations where you can separate variables is limited, but it includes an important class of problems. Once you are able to separate variables, the result is an ODE in each coordinate on an interval in the corresponding variable, which is why the region where you're solving needs to be a rectangular box in the chosen coordinate system. Sturm and Liouville did a nice job in the early 1800's of characterizing the ODEs coming out of separation of variables eigenvalue problems. These are the Sturm-Liouville eigenfunction equations with eigenvalue $lambda$ on an interval $[a,b]$: $ frac<1>left[fracleft(p(x)frac ight)+q(x) ight]=lambda f, cosalpha f(a)+sinalpha f'(a) = 0, coseta f(b)+sineta f'(b) = 0. $ The function $w$ determines a weight for the space $L^2_w(a,b)$ where the problem is properly posed as a selfadjoint one. The inner product on this space involves this weight function, and as given by $ (f,g)_w=int_^f(t)overlinew(t)dt. $ The function $p$ generally comes out of a scale factor associated with the coordinate change and is positive on $(a,b)$ and $q$ is a potential. Infinite and semi-infinite regions may occur (i.e., where $a=-infty$ and/or $b=infty$.) If $p$ vanishes at an endpoint of $(a,b)$, then the problem is singular, and there may or may not be some type of endpoint condition at the singular endpoint.

The cases grow in complexity with singularities, but the basic conclusion is this: you'll probably never have to worry about things not working the way you expect in regard to completeness. :)

Watch the video: Q153, Separable Differential Equation (July 2022).


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