### Video Transcript

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

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

This means that we can separate all the π¦ and dπ¦ 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 π¦ dπ¦ 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 π¦ with respect to π¦, note that one over three π¦ is equal to the constant one over three multiplied by one over π¦. 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 π¦ with respect to π¦ as one over three multiplied by the integral of one over π¦ with respect to π¦.

Now, recall that for nonzero π¦, the indefinite integral of one over π¦ with respect to π¦ is equal to the natural logarithm of the absolute value of π¦ plus π, where π 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 π¦ plus π one, where π one is the letter we have chosen to denote the constant of integration. Recall also that the indefinite integral of a constant π with respect to π₯ is equal to ππ₯ plus π, where π 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 π two, where π 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 π¦ is equal to three times π₯ plus three times π two minus three times π one. We can use the letter π to denote the constant formed by the difference three times π two minus three times π 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 π¦ is equal to the exponential function raised to the power of three π₯ plus π. We can split the right-hand side of the equation into π to the power of three π₯ multiplied by π to the power of π using the law of exponents relating to the product of powers. We can then use the letter capital π΄ to denote the constant π to the power of π. This completes the first step to solving a first order separable differential equation, which is to write the general solution in the form π¦ equals π of π₯, although in our case we have an absolute value sign around π¦.

The solution that we have found is called a general solution because it contains the unknown constant capital π΄. 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. Letβs use the initial condition π¦ equals two when π₯ equals three given to us in the question to determine the value of the constant π΄.

Substituting π¦ 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 π΄ multiplied by π to the power of three times three, which is just π΄ multiplied by π to the power of nine. Therefore, π΄ is equal to two divided by π to the power of nine, which we can rewrite as two multiplied by π 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 π¦ equal to two times π to the power of negative nine times π to the power of three π₯. Which we can rewrite as two times π 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 π 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 π to the power three π₯ minus nine, is strictly greater than zero for any real number π₯. This makes the absolute value sign on π¦ redundant. Hence, our particular solution is simply π¦ equals two times π to the power of three π₯ minus nine. Or π of π₯ equals two times π to the power of three π₯ minus nine, using the notation in the statement of the question.