Video: Solving a First-Order Separable Differential Equation with an Initial Condition

Let 𝑦′ = 1/2 𝑦 be the differential equation that approximately models the growth of the number of people infected with a certain disease. At time 𝑑 = 0 years, there are 30 people infected. Approximately, how many infected people will there be after 4 years?

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Video Transcript

Let 𝑦 prime equals a half 𝑦 be the differential equation that approximately models the growth of the number of people infected with a certain disease. At time 𝑑 equals zero years, there are 30 people infected. Approximately, how many infected people will there be after four years?

So we’ve been given the differential equation 𝑦 prime equals a half 𝑦 to model the spread of this particular disease. Now, another way to write this would be as d𝑦 by d𝑑 equals a half 𝑦. The rate of change of 𝑦 with respect to time equals one-half 𝑦. And if we express the differential equation in this way, then we’ll perhaps see more easily that it is in fact a separable differential equation. Now, whilst d𝑦 by d𝑑 is not a fraction, we do treat it a little like one. And so it is equivalent to say that one over 𝑦 d𝑦 is equal to a half d𝑑. And so we’ve separated the variables. We have all the 𝑦’s on the left-hand side and all the terms involving 𝑑 on the right-hand side.

We’re then going to integrate both sides of this equation. Integrating the left-hand side, we recall that the integral of one ever 𝑦 with respect to 𝑦 is equal to the natural logarithm of the absolute value of 𝑦. And we can include a constant of integration, 𝑐 one. On the right-hand side, the integral of a constant with respect to 𝑑, in this case a half, is a half 𝑑. And we can include a second constant of integration, 𝑐 two. Now, in fact, by subtracting 𝑐 one from each side, we can combine our two constants of integration into a single constant of integration, which we’ll call 𝑐. And we have that the natural logarithm of the absolute value of 𝑦 is equal to a half 𝑑 plus 𝑐.

Now, remember here that 𝑦 represents the number of people infected with this disease. And therefore, by definition, 𝑦 must be greater than or equal to zero. Therefore, the absolute value of 𝑦 is just 𝑦 itself. And so we can remove these absolute value signs, giving that the natural logarithm of 𝑦 is equal to a half 𝑑 plus 𝑐. The next step in solving this equation to give 𝑦 in terms of 𝑑 is to take 𝑒 to the power of each side, knowing that this will cancel out the natural logarithm on the left-hand side. When we do so, we get 𝑦 is equal to 𝑒 to the power of a half 𝑑 plus 𝑐.

Next, we simplify the right-hand side using one of our laws of exponents, which tells us that π‘₯ to the power of π‘Ž multiplied by π‘₯ to the power of 𝑏 is equal to the π‘₯ to the power of π‘Ž plus 𝑏. So, using this rule in reverse, we can say that, on the right-hand side, we have 𝑒 to the power of 𝑐 multiplied by 𝑒 to the power of a half 𝑑. Now, 𝑐, remember, was a constant. And so 𝑒 to the power of 𝑐 is just a different constant, which we can call 𝐴. And so we have the general solution to this differential equation. 𝑦 is equal to 𝐴 multiplied by 𝑒 to the power of one-half 𝑑 for some constant 𝐴.

We’re asked to determine how many infected people there will be after four years. So that’s the value of 𝑦 when 𝑑 equals four. But in order to do this, we first need to determine the value of the constant 𝐴. We can do this by considering the other information we’re given in the question, which is an initial condition. We know that, at time 𝑑 equals zero, there are 30 people infected. So when 𝑑 equals zero, 𝑦 equals 30.

Substituting 30 for 𝑦 and zero for 𝑑 gives 30 equals 𝐴 multiplied by 𝑒 to the power of a half multiplied by zero. Now, a half multiplied by zero is just zero. And 𝑒 to the power of zero is one. So we have 30 is equal to 𝐴 multiplied by one, which simplifies to 30 equals 𝐴. We have therefore found the value of this constant. The particular solution to this differential equation then is 𝑦 equals 30𝑒 to the power of one-half 𝑑.

Finally, we need to substitute 𝑑 equals four to determine the number of infected people after four years. Doing so gives 𝑦 equals 30𝑒 to the power of four over two. That’s two. So we have 𝑦 equals 30𝑒 squared. Now, we’re using a continuous function here to model a discrete variable. The number of people must be an integer value. So we could evaluate 30𝑒 squared. But we’d then need to round this answer to the nearest integer. We can say then that the number of infected people after four years will be an integer or whole number close to this value of 30𝑒 squared.

So in this question, we separated the variables in our differential equation, integrated both sides to find an expression for 𝑦 in terms of 𝑑. We then used the initial condition to find the value of our constant 𝐴 and, finally, substituted 𝑑 equals four to determine approximately how many infected people there will be after four years. And we found that this number will be a whole number close to 30𝑒 squared.

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