Video: Solving a First Order Separable Differential Equation

If 𝑓′(π‘₯) = 3𝑓(π‘₯) and 𝑓(3) = 2, find an expression for 𝑓(π‘₯).

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Video Transcript

If 𝑓 prime of π‘₯ equals three times 𝑓 of π‘₯ and 𝑓 evaluated at three equals two, find an expression for 𝑓 of π‘₯.

Let’s right 𝑓 prime of π‘₯ as d𝑦 by dπ‘₯ and 𝑓 of π‘₯ as 𝑦. Then, the differential equation 𝑓 prime of π‘₯ equals three times 𝑓 of π‘₯ translates to d𝑦 by dπ‘₯ equals three times 𝑦. The initial condition 𝑓 of three equals two translates to 𝑦 equals two when π‘₯ equals three. We are required to find the particular solution 𝑦 enclosed 𝑓 of π‘₯ that results from this differential equation and initial condition. Note that the differential equation d𝑦 by dπ‘₯ equals three times 𝑦 is a separable differential equation.

This means that we can separate all the 𝑦 and d𝑦 terms to one side of the equation and all the π‘₯ and dπ‘₯ terms to the other side of the equation. Doing so, we obtain one over three 𝑦 d𝑦 equals one dπ‘₯. Separating the variables like so is the first step to solving a first order separable differential equation. The second step is to integrate both sides of the equation. To integrate one over three 𝑦 with respect to 𝑦, note that one over three 𝑦 is equal to the constant one over three multiplied by one over 𝑦. Therefore, using the fact that a constant factor inside of an integral can be pulled outside the integral, we can rewrite the integral of one over three 𝑦 with respect to 𝑦 as one over three multiplied by the integral of one over 𝑦 with respect to 𝑦.

Now, recall that for nonzero 𝑦, the indefinite integral of one over 𝑦 with respect to 𝑦 is equal to the natural logarithm of the absolute value of 𝑦 plus 𝑐, where 𝑐 is the constant of integration. So, the left-hand side of our equation becomes one over three multiplied by the natural logarithm of the absolute value of 𝑦 plus 𝑐 one, where 𝑐 one is the letter we have chosen to denote the constant of integration. Recall also that the indefinite integral of a constant π‘Ž with respect to π‘₯ is equal to π‘Žπ‘₯ plus 𝑐, where 𝑐 is the constant of integration.

So, the indefinite integral of the constant one with respect to π‘₯ is one times π‘₯, which is just π‘₯, plus 𝑐 two, where 𝑐 two is the letter we have chosen to denote the constant of integration on the right-hand side of the equation. Collecting the constants of integration together on the right-hand side of the equation and then multiplying both hand sides of the equation by three, we obtain that the natural logarithm of the absolute value 𝑦 is equal to three times π‘₯ plus three times 𝑐 two minus three times 𝑐 one. We can use the letter 𝑐 to denote the constant formed by the difference three times 𝑐 two minus three times 𝑐 one.

Applying the exponential function, which is the inverse function, to the natural logarithms function to both sides of the equation, we obtain that the absolute value of 𝑦 is equal to the exponential function raised to the power of three π‘₯ plus 𝑐. We can split the right-hand side of the equation into 𝑒 to the power of three π‘₯ multiplied by 𝑒 to the power of 𝑐 using the law of exponents relating to the product of powers. We can then use the letter capital 𝐴 to denote the constant 𝑒 to the power of 𝑐. This completes the first step to solving a first order separable differential equation, which is to write the general solution in the form 𝑦 equals 𝑓 of π‘₯, although in our case we have an absolute value sign around 𝑦.

The solution that we have found is called a general solution because it contains the unknown constant capital 𝐴. The fourth, and final, step to solving a first order separable differential equation is to use the initial conditions, if given, to find a particular solution. The particular solution should not contain any undetermined constants. Let’s use the initial condition 𝑦 equals two when π‘₯ equals three given to us in the question to determine the value of the constant 𝐴.

Substituting 𝑦 equals two into the left-hand side of the general solution, we obtain the absolute value of two, which is just two. Substituting π‘₯ equals three into the right-hand side of the general equation, we obtain 𝐴 multiplied by 𝑒 to the power of three times three, which is just 𝐴 multiplied by 𝑒 to the power of nine. Therefore, 𝐴 is equal to two divided by 𝑒 to the power of nine, which we can rewrite as two multiplied by 𝑒 to the power of negative nine.

So, the particular solution to the given differential equation with the given initial condition is the absolute value of 𝑦 equal to two times 𝑒 to the power of negative nine times 𝑒 to the power of three π‘₯. Which we can rewrite as two times 𝑒 to the power of three π‘₯ minus nine using the law of exponents relating to the product of powers. Now, note that two is strictly greater than zero. Also note that 𝑒 to the power of three π‘₯ minus nine is strictly greater than zero for all real numbers π‘₯ as the exponential function raised to any real power always results in a positive number.

Therefore, their product, two times 𝑒 to the power three π‘₯ minus nine, is strictly greater than zero for any real number π‘₯. This makes the absolute value sign on 𝑦 redundant. Hence, our particular solution is simply 𝑦 equals two times 𝑒 to the power of three π‘₯ minus nine. Or 𝑓 of π‘₯ equals two times 𝑒 to the power of three π‘₯ minus nine, using the notation in the statement of the question.

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