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Lesson Explainer: Separable Differential Equations Mathematics • Higher Education

In this explainer, we will learn how to identify and solve separable differential equations.

A differential equation is an equation that relates functions to their derivativesβ€”that is, an equation expressing a relationship between functions and their rates of change. As such, differential equations are extremely prevalent in engineering and physicsβ€”many of the fundamental laws of physics are expressed in terms of differential equations. A differential equation may involve multiple functions, their derivatives, their second and higher derivatives, and even their partial derivatives. In this explainer, we will be looking at only the simplest case of differential equations, which is that of a single function in a single variable and its first derivative. Such equations are called first-order ordinary differential equations.

Consider the equation dd𝑦π‘₯=2π‘₯.

This is an example of a first-order ordinary differential equation. It has a particular solution, which is the function 𝑦=π‘₯, because the first derivative of 𝑦=π‘₯ is indeed 2π‘₯. However, this is not the only solution to the equation. Indeed, any function of the form 𝑦=π‘₯+𝐢, where 𝐢 is a constant, satisfies the differential equation dd𝑦π‘₯=2π‘₯. The equation 𝑦=π‘₯+𝐢 is called the general solution to the differential equation. It represents a family of solutions, parameterized by πΆβˆˆβ„.

Since solving a differential equation involves β€œgetting rid” of a derivative, it should come as no surprise that the basic tool for solving differential equations is integration. We will be looking at a class of differential equations for which the application of integration is relatively straightforward. A separable first-order ordinary differential equation is an equation of the form dd𝑦π‘₯=𝑔(π‘₯)β„Ž(𝑦), in which 𝑔(π‘₯) is an expression purely in π‘₯ and β„Ž(𝑦) is an expression purely in 𝑦. First, note that there is an obvious family of solutions to this equation in the form of any constant function 𝑦=π‘Ž (implying that dd𝑦π‘₯=0), such that β„Ž(π‘Ž)=0 (which makes the differential equation true). These solutions are called equilibrium solutions to the differential equation.

To find the nonequilibrium solutions, we suppose that β„Ž(𝑦)β‰ 0 and divide both sides by β„Ž(𝑦), β€œseparating the variables” to get the equation 1β„Ž(𝑦)𝑦π‘₯=𝑔(π‘₯).dd

The next step in solving the differential equation is to integrate both sides of this equation with respect to π‘₯: ο„Έ1β„Ž(𝑦)𝑦π‘₯π‘₯=𝑔(π‘₯)π‘₯.dddd

Recall that the rules of integration by substitution tell us that ο„Έ1β„Ž(𝑦)𝑦π‘₯π‘₯=ο„Έ1β„Ž(𝑦)𝑦.dddd

Therefore, we have arrived at ο„Έ1β„Ž(𝑦)𝑦=𝑔(π‘₯)π‘₯.dd

This procedure holds in general for any separable differential equation dd𝑦π‘₯=𝑔(π‘₯)β„Ž(𝑦). Let us summarize the situation in the following fundamental fact.

Rule: General Solution of Separable Differential Equations

If the equation dd𝑦π‘₯=𝑔(π‘₯)β„Ž(𝑦) is true and β„Ž(𝑦)β‰ 0, then the equation ο„Έ1β„Ž(𝑦)𝑦=𝑔(π‘₯)π‘₯dd is true.

We can now solve the equation by calculating the integrals. Let us look at an example.

Consider the equation dd𝑦π‘₯=π‘₯𝑦.

This is a separable first-order ordinary differential equation, with 𝑔(π‘₯)=π‘₯ and β„Ž(𝑦)=𝑦. Observe that 𝑦=0 is a constant solution to the equation β„Ž(𝑦)=0. We note this down as an equilibrium solution to dd𝑦π‘₯=π‘₯𝑦 and assume that β„Ž(𝑦)β‰ 0. Next, we divide both sides by β„Ž(𝑦)=𝑦 to get 1𝑦𝑦π‘₯=π‘₯.dd

By our general rule above, this yields the integral equation ο„Έ1𝑦𝑦=ο„Έπ‘₯π‘₯,dd which we can calculate. Not forgetting the constants of integration, since these are indefinite integrals, we have ln|𝑦|+𝐢=12π‘₯+𝐢.

Because 𝐢 and 𝐢 are both constants, we are going to subtract 𝐢 from both sides to form a new constant 𝐢=πΆβˆ’πΆοŠ¨οŠ§: ln|𝑦|=12π‘₯+𝐢.

This is a general solution to the differential equation dd𝑦π‘₯=π‘₯𝑦. However, it is good to rearrange our solution into the form 𝑦=𝐹(π‘₯), where 𝐹(π‘₯) is some function of π‘₯, if we can. So, we apply the exponential map to both sides: |𝑦|=𝑒=𝑒𝑒.οŽ οŽ‘οŽ‘οŽ οŽ‘οŽ‘ο—οŠ°οŒ’ο—οŒ’

Notice that the absolute value around 𝑦 indicates that we have two cases; namely, 𝑦=𝑒𝑒,𝑦β‰₯0,𝑦=βˆ’π‘’π‘’,𝑦<0.οŽ οŽ‘οŽ‘οŽ οŽ‘οŽ‘ο—οŒ’ο—οŒ’

We combine these two cases and eliminate the absolute value by defining a new constant 𝐴=𝑒𝑦β‰₯0,βˆ’π‘’π‘¦<0,ifif giving us the general solution 𝑦=𝐴𝑒.οŽ οŽ‘οŽ‘ο—

So, the differential equation dd𝑦π‘₯=π‘₯𝑦 has the particular equilibrium solution 𝑦=0 and the general solution family 𝑦=π΄π‘’οŽ οŽ‘οŽ‘ο— parameterized by π΄βˆˆβ„.

Let us look now at a slightly more complicated example.

Example 1: Finding the General Solution to a Separable Differential Equation

Solve the differential equation dd𝑦π‘₯=βˆ’5π‘₯βˆšπ‘¦.

Answer

We have a general procedure for solving such separable differential equations, which is as follows:

  1. We have a separable equation in the form dd𝑦π‘₯=𝑔(π‘₯)β„Ž(𝑦), so we first check for any equilibrium solutions in the form of constant solutions to the equation β„Ž(𝑦)=0.
  2. Next, we suppose β„Ž(𝑦)β‰ 0 and divide by β„Ž(𝑦): 1β„Ž(𝑦)𝑦π‘₯=𝑔(π‘₯).dd
  3. Next, we integrate this equation, which gives ο„Έ1β„Ž(𝑦)𝑦=𝑔(π‘₯)π‘₯.dd
  4. Now, we compute the integral on both sides, not forgetting the constants of integration.
  5. Finally, we rearrange the resulting equation into the form 𝑦=𝐹(π‘₯), if possible.

We have an equation of the form dd𝑦π‘₯=𝑔(π‘₯)β„Ž(𝑦), with 𝑔(π‘₯)=βˆ’5π‘₯ and β„Ž(𝑦)=βˆšπ‘¦. Before proceeding, we check for equilibrium solutions in the form of constant solutions to β„Ž(𝑦)=0 and see that we have one in the form of 𝑦=0. We note this down for later and now we suppose 𝑦≠0. We can now divide both sides of the equation by β„Ž(𝑦)=βˆšπ‘¦: 1βˆšπ‘¦π‘¦π‘₯=βˆ’5π‘₯.dd

Next, we form the integral equation 𝑦𝑦=βˆ’5ο„Έπ‘₯π‘₯,dd making the change of notation 1βˆšπ‘¦=π‘¦οŠ±οŽ οŽ‘ for convenience. Now, we integrate 2𝑦+𝐢=βˆ’52π‘₯+𝐢, combine constants on the right-hand side to get 2𝑦=βˆ’52π‘₯+𝐢, and divide by 2 to get 𝑦=βˆ’54π‘₯+𝐢.

Since the question did not specify in what form to give our solution, we can write 𝑦=βˆ’54π‘₯+𝐢 as the general solution to the differential equation dd𝑦π‘₯=βˆ’5π‘₯βˆšπ‘¦ and 𝑦=0 as the equilibrium solution. We can also write 𝑦=ο€Όβˆ’54π‘₯+𝐢=2516π‘₯βˆ’52𝐢π‘₯+𝐢οŠͺ as an acceptable form for the general solution.

In the previous example, we were given a separable differential equation in the form dd𝑦π‘₯=𝑔(π‘₯)β„Ž(𝑦). Not every separable differential looks like this; some rearrangement may first be required, as in the next example.

Example 2: Finding the General Solution to a Separable Differential Equation

Solve the differential equation dd𝑦π‘₯+𝑦=1.

Answer

Here, we have a separable first-order ordinary differential equation dd𝑦π‘₯+𝑦=1 that is not presented to us in standard separable form dd𝑦π‘₯=𝑔(π‘₯)β„Ž(𝑦). The procedure for solving such equations is as follows:

  1. First, we rearrange the equation into the form dd𝑦π‘₯=𝑔(π‘₯)β„Ž(𝑦) and check for any equilibrium solutions in the form of constant solutions to the equation β„Ž(𝑦)=0.
  2. Now, we suppose β„Ž(𝑦)β‰ 0 and divide by β„Ž(𝑦): 1β„Ž(𝑦)𝑦π‘₯=𝑔(π‘₯).dd
  3. Next, we integrate this equation, which gives ο„Έ1β„Ž(𝑦)𝑦=𝑔(π‘₯)π‘₯.dd
  4. Now, we compute the integral on both sides, not forgetting the constants of integration.
  5. Finally, we rearrange the resulting equation into the form 𝑦=𝐹(π‘₯), if possible.

Our first step, then, is to rearrange dd𝑦π‘₯+𝑦=1 into the form dd𝑦π‘₯=𝑔(π‘₯)β„Ž(𝑦). So, let us subtract 𝑦 from both sides of the equation to get dd𝑦π‘₯=1βˆ’π‘¦.

Here, we have β„Ž(𝑦)=(1βˆ’π‘¦) and 𝑔(π‘₯)=1. We can see that 𝑦=1 solves β„Ž(𝑦)=0 and is an equilibrium solution to dd𝑦π‘₯+𝑦=1.

The next step is to suppose 𝑦≠1 and divide by β„Ž(𝑦)=(1βˆ’π‘¦): 11βˆ’π‘¦π‘¦π‘₯=1.dd

Now, we form the integral equation ο„Έ11βˆ’π‘¦π‘¦=ο„Έ1π‘₯dd and compute the indefinite integrals, not forgetting the constants of integration: βˆ’|1βˆ’π‘¦|+𝐢=π‘₯+𝐢.ln

We want to rearrange this into the form 𝑦=𝐹(π‘₯), where 𝐹(π‘₯) is some function of π‘₯. First, we multiply everything by βˆ’1 and define a new constant 𝐢=πΆβˆ’πΆοŠ§οŠ¨: ln|1βˆ’π‘¦|=βˆ’π‘₯+𝐢.

Now, we apply the exponential to both sides to get 1βˆ’π‘¦=𝑒,οŠ±ο—οŠ°οŒ’ define a new new constant 𝐴=π‘’οŒ’ to get 1βˆ’π‘¦=𝐴𝑒,οŠ±ο— subtract 1 to get βˆ’π‘¦=π΄π‘’βˆ’1,οŠ±ο— and finally, multiply by βˆ’1 again, observing that 𝐴, being some unspecified number, may as well β€œabsorb” this βˆ’1, remaining an unspecified number: 𝑦=𝐴𝑒+1.οŠ±ο—

Thus, 𝑦=𝐴𝑒+1οŠ±ο— is our general solution to the differential equation dd𝑦π‘₯+𝑦=1 and 𝑦=1 is the equilibrium solution.

Sometimes, we encounter a separable differential equation presented in such a way that it is more convenient to rearrange directly into the form 𝐻(𝑦)𝑦π‘₯=𝐺(π‘₯)dd rather than first into the form dd𝑦π‘₯=𝑔(π‘₯)β„Ž(𝑦). The next examples demonstrate this.

Example 3: Finding the General Solution to a Separable Differential Equation

Solve the following differential equation: (π‘’βˆ’5)𝑦′=2+(π‘₯)cos.

Answer

In this question, we have a separable differential equation given to us in a form in which the variables are already separated; that is, 𝐻(𝑦)𝑦′=𝐺(π‘₯), where 𝐻(𝑦)=(π‘’βˆ’5) and 𝐺(π‘₯)=2+(π‘₯)cos. The notation 𝑦′ is simply the Lagrangian notation for the derivative 𝑦′=𝑦π‘₯dd. We proceed straight to integrating: ο„Έ(π‘’βˆ’5)𝑦=ο„Έ(2+(π‘₯))π‘₯ο„Έπ‘’π‘¦βˆ’5ο„Έ1𝑦=ο„Έ2π‘₯+ο„Έ(π‘₯)π‘₯π‘’βˆ’5𝑦+𝐢=2π‘₯+(π‘₯)+𝐢.dcosddddcosdsin

Combining constants on the right-hand side, we have π‘’βˆ’5𝑦=2π‘₯+(π‘₯)+𝐢sin as the general solution to the differential equation (π‘’βˆ’5)𝑦′=2+(π‘₯)cos.

So far, we have dealt only with general solutions to differential equationsβ€”that is, with families of solutions parameterized by a constant 𝐢. However, it is possible to deduce the value of 𝐢 if we are given a pair of values that the solution must satisfy. These given values are called boundary or initial conditions, and the resulting unique solution satisfying these conditions is called a specific solution to the differential equation.

Consider again our example dd𝑦π‘₯=π‘₯𝑦 with the general solution 𝑦=𝐴𝑒,οŽ οŽ‘οŽ‘ο— where 𝐴 is a constant. Suppose now we are given the boundary conditions that 𝑦=5 when π‘₯=0. We simply substitute these values into our general solution to find the value of the constant 𝐴: 5=𝐴𝑒5=𝐴𝑒𝐴=5.

So, the particular solution to the differential equation dd𝑦π‘₯=π‘₯𝑦, given that 𝑦=5 when π‘₯=0, is 𝑦=5π‘’οŽ οŽ‘οŽ‘ο—.

Example 4: Finding the Particular Solution to a Separable Differential Equation

Solve the following differential equation, using the given boundary conditions to find a particular solution: cossecddcscsec(π‘₯)(𝑦)𝑦π‘₯=(𝑦)(π‘₯),π‘₯=0,𝑦=πœ‹4.

Answer

To find the particular solution to a separable differential equation, such as the one given in the question, we proceed just as we would to find its general solution and then substitute in the given boundary conditions at the end, following these steps:

  1. First, we rearrange the equation directly into the form 𝐻(𝑦)𝑦π‘₯=𝐺(π‘₯).dd
  2. Next, we integrate this equation, which gives 𝐻(𝑦)𝑦=𝐺(π‘₯)π‘₯.dd
  3. Now, we compute both integrals, not forgetting the constants of integration.
  4. After rearranging, this will yield a general solution to the differential equation in the form 𝑃(𝑦)=𝑄(π‘₯), where 𝑃(𝑦) is an expression in 𝑦 and 𝑄(π‘₯) is an expression in π‘₯ containing some constant whose value is to be determined. We substitute the given boundary conditions into this equation to find the value of the constant.

First, we want to rearrange the equation cossecddcscsec(π‘₯)(𝑦)𝑦π‘₯=(𝑦)(π‘₯) into the form 𝐻(𝑦)𝑦π‘₯=𝐺(π‘₯)ddβ€”that is, with all expressions in 𝑦 on the left-hand side and all expressions in π‘₯ on the right-hand side. Observe first that the presence of sec(π‘₯) on the right-hand side means that this equation is not defined when cos(π‘₯)=0β€”that is, when π‘₯=(2𝑛+1)πœ‹2 and π‘›βˆˆβ„€, so we exclude these cases and divide both sides by coscsc(π‘₯)(𝑦) to get seccscddseccos(𝑦)(𝑦)𝑦π‘₯=(π‘₯)(π‘₯) and rewrite things a little, using the identities seccos=1 and cscsin=1 to get tanddsec(𝑦)𝑦π‘₯=(π‘₯).

We are now ready to integrate (combining constants of integration on the right-hand side): ο„Έ(𝑦)𝑦=ο„Έ(π‘₯)π‘₯.tandsecd

To compute the left-hand integral, we use the identity tansec(𝑦)≑(𝑦)βˆ’1, so ο„Έ(𝑦)𝑦=ο„Έο€Ή(𝑦)βˆ’1𝑦=ο„Έ(𝑦)π‘¦βˆ’ο„Έ1𝑦,tandsecdsecdd and we recall that ddtansec𝑦(𝑦)=(𝑦), giving us tantan(𝑦)βˆ’π‘¦=(π‘₯)+𝐢 as a general solution to our differential equation.

Finally, we substitute the boundary conditions π‘₯=0 and 𝑦=πœ‹4 into the general solution tantan(𝑦)βˆ’π‘¦=(π‘₯)+𝐢 to find a value for 𝐢, tantanο€»πœ‹4ο‡βˆ’πœ‹4=(0)+𝐢1βˆ’πœ‹4=𝐢, and the particular solution, tantan(𝑦)βˆ’π‘¦=(π‘₯)+1βˆ’πœ‹4.

Finally, we look at an example that requires a little more work to solve.

Example 5: Finding the Particular Solution to a Separable Differential Equation

By expressing 9π‘₯βˆ’17(4π‘₯βˆ’7)(π‘₯βˆ’3) in partial fractions, write down the particular solution to the differential equation (4π‘₯βˆ’7)(π‘₯βˆ’3)𝑦π‘₯=(9π‘₯βˆ’17)𝑦,dd satisfying the condition 𝑦=5 when π‘₯=2. Give your solution in the form 𝑦=𝑓(π‘₯).

Answer

We solve this problem in the following steps:

  1. First, we rearrange the equation into the form dd𝑦π‘₯=𝑔(π‘₯)β„Ž(𝑦) and check for any equilibrium solutions in the form of constant solutions to the equation β„Ž(𝑦)=0.
  2. Now, we suppose β„Ž(𝑦)β‰ 0 and divide by β„Ž(𝑦): 1β„Ž(𝑦)𝑦π‘₯=𝑔(π‘₯).dd
  3. Next, we integrate this equation, which gives ο„Έ1β„Ž(𝑦)𝑦=𝑔(π‘₯)π‘₯.dd
  4. Now, we compute both integrals, not forgetting the constants of integration.
  5. We rearrange the resulting equation into the form 𝑦=𝐹(π‘₯), where 𝐹(π‘₯) is an expression in π‘₯ containing some unknown constant to be determined.
  6. Finally, we substitute in the given boundary conditions to find the value of the unknown constant and write down a particular solution to the equation.

Observe first that we have an equilibrium solution to (4π‘₯βˆ’7)(π‘₯βˆ’3)𝑦π‘₯=(9π‘₯βˆ’17)𝑦dd at 𝑦=0. The particular values π‘₯=3 and π‘₯=74 that make the left-hand side 0 also force 𝑦=0. Thus, excluding 𝑦=0 also excludes these values of π‘₯, and so we can rearrange the equation (4π‘₯βˆ’7)(π‘₯βˆ’3)𝑦π‘₯=(9π‘₯βˆ’17)𝑦dd into the form dd𝑦π‘₯=𝑔(π‘₯)β„Ž(𝑦) by dividing by (4π‘₯βˆ’7)(π‘₯βˆ’3): dd𝑦π‘₯=(9π‘₯βˆ’17)(4π‘₯βˆ’7)(π‘₯βˆ’3)𝑦.

Here, 𝑔(π‘₯)=(9π‘₯βˆ’17)(4π‘₯βˆ’7)(π‘₯βˆ’3) and β„Ž(𝑦)=𝑦. Since we have excluded the equilibrium solution 𝑦=0, we can divide by β„Ž(𝑦)=𝑦 for 1𝑦𝑦π‘₯=(9π‘₯βˆ’17)(4π‘₯βˆ’7)(π‘₯βˆ’3).dd

The next step is to integrate: ο„Έ1𝑦𝑦=ο„Έ(9π‘₯βˆ’17)(4π‘₯βˆ’7)(π‘₯βˆ’3)π‘₯.dd

The left-hand side is straightforward enough, ο„Έ1𝑦𝑦=|𝑦|+𝐢dln. In order to tackle the right-hand side, we need to express (9π‘₯βˆ’17)(4π‘₯βˆ’7)(π‘₯βˆ’3) in partial fractions. So, we put (9π‘₯βˆ’17)(4π‘₯βˆ’7)(π‘₯βˆ’3)=𝐴4π‘₯βˆ’7+𝐡π‘₯βˆ’3.

We multiply 𝐴4π‘₯βˆ’7 by π‘₯βˆ’3π‘₯βˆ’3 and 𝐡π‘₯βˆ’3 by 4π‘₯βˆ’74π‘₯βˆ’7 so that we have a common denominator: (9π‘₯βˆ’17)(4π‘₯βˆ’7)(π‘₯βˆ’3)=𝐴4π‘₯βˆ’7+𝐡π‘₯βˆ’3=𝐴(π‘₯βˆ’3)(4π‘₯βˆ’7)(π‘₯βˆ’3)+𝐡(4π‘₯βˆ’7)(π‘₯βˆ’3)(4π‘₯βˆ’7)=𝐴(π‘₯βˆ’3)+𝐡(4π‘₯βˆ’7)(4π‘₯βˆ’7)(π‘₯βˆ’3)=𝐴π‘₯βˆ’3𝐴+4𝐡π‘₯βˆ’7𝐡(4π‘₯βˆ’7)(π‘₯βˆ’3).

Looking at the numerators, we have the equation 9π‘₯βˆ’17=𝐴π‘₯βˆ’3𝐴+4𝐡π‘₯βˆ’7𝐡.

Equating π‘₯-coefficients gives 9=𝐴+4𝐡, and equating constant terms gives βˆ’17=βˆ’3π΄βˆ’7𝐡.

We can solve this pair of simultaneous equations by elimination. First, we multiply 9=𝐴+4𝐡 by 3 for 27=3𝐴+12𝐡 and then add this to βˆ’17=βˆ’3π΄βˆ’7𝐡: βˆ’17+27=βˆ’3π΄βˆ’7𝐡+(3𝐴+12𝐡)10=5𝐡𝐡=2.

Now, we substitute 𝐡=2 back into 9=𝐴+4𝐡 so that 9=𝐴+8 and 𝐴=1. Thus, we have (9π‘₯βˆ’17)(4π‘₯βˆ’7)(π‘₯βˆ’3)=14π‘₯βˆ’7+2π‘₯βˆ’3.

We are now ready to integrate: ο„Έ1𝑦𝑦=ο„Έ(9π‘₯βˆ’17)(4π‘₯βˆ’7)(π‘₯βˆ’3)π‘₯|𝑦|+𝐢=ο„Έ14π‘₯βˆ’7+2π‘₯βˆ’3π‘₯=ο„Έ14π‘₯βˆ’7π‘₯+ο„Έ2π‘₯βˆ’3π‘₯=14|4π‘₯βˆ’7|+2|π‘₯βˆ’3|+𝐢.ddlndddlnln

Consolidating constants on the right, we have lnlnln|𝑦|=14|4π‘₯βˆ’7|+2|π‘₯βˆ’3|+𝐢, which is a general solution to the differential equation (4π‘₯βˆ’7)(π‘₯βˆ’3)𝑦π‘₯=(9π‘₯βˆ’17)𝑦dd. In order to find a particular solution, we need to calculate the value of the constant. Before doing this, we rearrange our general solution into the form 𝑦=𝐹(π‘₯) by applying the exponential: |𝑦|=(|4π‘₯βˆ’7|)(|π‘₯βˆ’3|)𝑒.

Observe that (π‘₯βˆ’3)=(3βˆ’π‘₯), so (|π‘₯βˆ’3|)=(π‘₯βˆ’3). The remaining pair of expressions inside modulus bars gives us four cases to consider: 𝑦=⎧βŽͺβŽͺβŽͺ⎨βŽͺβŽͺβŽͺ⎩(4π‘₯βˆ’7)(π‘₯βˆ’3)𝑒𝑦β‰₯0,π‘₯>74,(7βˆ’4π‘₯)(π‘₯βˆ’3)𝑒𝑦β‰₯0,π‘₯<74,βˆ’(4π‘₯βˆ’7)(π‘₯βˆ’3)𝑒𝑦<0,π‘₯>74,βˆ’(7βˆ’4π‘₯)(π‘₯βˆ’3)𝑒𝑦<0,π‘₯<74.ifififif

We define a new constant 𝐴=𝑒𝑦β‰₯0,βˆ’π‘’π‘¦<0,ifif leaving us with the general solution 𝑦=⎧⎨⎩𝐴(4π‘₯βˆ’7)(π‘₯βˆ’3)π‘₯>74,𝐴(7βˆ’4π‘₯)(π‘₯βˆ’3)π‘₯<74.ifif

The picture shows an example of the two β€œbranches” of the general solution, here, taken with the parameter 𝐴=1. Observe that the function 𝑦 can be defined at π‘₯=74 to be 𝑦=0, but it is certainly not differentiable there. Thus, π‘₯=74 is not part of the domain of the general solution.

The last step in finding a particular solution is to use the boundary conditions 𝑦=5 and π‘₯=2. Since 2>74, we are dealing now with the function 𝑦=𝐴(4π‘₯βˆ’7)(π‘₯βˆ’3). Substituting the boundary conditions into this general solution, we have 5=𝐴(4Γ—2βˆ’7)(2βˆ’3)=𝐴(8βˆ’7)(βˆ’1)=𝐴×1Γ—1=𝐴.

We finally substitute the value 𝐴=5 into the general solution for the particular solution 𝑦=5(4π‘₯βˆ’7)(π‘₯βˆ’3). Thus, we have 2 solutions: this one and the equilibrium 𝑦=0.

In this picture, the green curve is the solution 𝑦=5(4π‘₯βˆ’7)(π‘₯βˆ’3) and the red line is the equilibrium solution 𝑦=0. Observe that the solution 𝑦=5(4π‘₯βˆ’7)(π‘₯βˆ’3) is not defined for π‘₯<74 and is not differentiable at the point π‘₯=74. It is therefore valid on the open interval 74,βˆžο”. The domain of the equilibrium solution is all of ℝ.

Let us finish by recapping a few important concepts from this explainer.

Key Points

  • We recognize separable differential equations as differential equations that can be rearranged into the form dd𝑦π‘₯=𝑔(π‘₯)β„Ž(𝑦).
  • We understand that constant solutions to the equation β„Ž(𝑦)=0 represent equilibrium solutions to the differential equation.
  • In order to find the general solution to a separable differential equation dd𝑦π‘₯=𝑔(π‘₯)β„Ž(𝑦), we suppose β„Ž(𝑦)β‰ 0 and divide by β„Ž(𝑦) to get 1β„Ž(𝑦)𝑦π‘₯=𝑔(π‘₯)dd and integrate both sides of ο„Έ1β„Ž(𝑦)𝑦=𝑔(π‘₯)π‘₯.dd
  • The general solution to such a differential equation represents an infinite family of solutions, parameterized by a constant 𝐢.
  • In order to find a particular solution given boundary conditions π‘₯=π‘₯ and 𝑦=π‘¦οŠ¦, we substitute these values into the general solution to find the value of the constant 𝐢.

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