Video Transcript
Suppose that d𝑦 by d𝑥 is equal to 𝑦 to the power of four and 𝑦 is equal to negative one when 𝑥 is equal to one-third. What will the value of 𝑦 be when 𝑥 is equal to nine?
Now, we’ve been given a first-order differential equation, which is d𝑦 by d𝑥 is equal to 𝑦 to the power of four. In order to find the value of 𝑦 when 𝑥 is equal to nine, we’re going to need to solve this differential equation. Then find a particular solution when 𝑦 is equal to negative one and 𝑥 is equal to one-third. And then use the equation in order to find the value of 𝑦. Our differential equation is of the form d𝑦 by d𝑥 is equal to 𝑓 of 𝑥 multiplied by 𝑔 of 𝑦, where 𝑓 of 𝑥 is equal to one and 𝑔 of 𝑦 is equal to 𝑦 to the power of four.
Now this type of first-order differential equation is called a separable differential equation since we can separate the variables within the equation in order to integrate. We will obtain an equation of the form one over 𝑔 of 𝑦 d𝑦 is equal to 𝑓 of 𝑥 d𝑥. And we can simply integrate both sides in order to solve it. In our case, we have d𝑦 by d𝑥 is equal to 𝑦 to the power of four. Separating the variables gives us that one over 𝑦 to the power of four d𝑦 is equal to one d𝑥. And now, we simply integrate both sides of the equation.
On the right-hand side, we’re simply integrating one with respect to 𝑥. This will give us 𝑥 plus some constant of integration 𝐴. In order to find the integral on the left, we can rewrite one over 𝑦 to the power of four as 𝑦 to the power of negative four. In order to integrate 𝑦 to the power of negative four with respect to 𝑦, we simply need to increase the power of 𝑦 by one and divide by the new power. If we increase negative four by one, we reach negative three, giving us 𝑦 to the power of negative three. And then we need to divide by the new power. So that’s negative three. We mustn’t forget to add on a constant of integration, which we can call 𝐵.
Now we have found that negative one-third 𝑦 to the power of negative three plus 𝐵 is equal to 𝑥 plus 𝐴. Now we can simply subtract 𝐵 from both sides. And now since we have a constant of 𝐴 minus 𝐵 on the right-hand side, we can combine these two together to a third constant, which we can call 𝐶. Next, we can multiply both sides of the equation by negative three. And we can rewrite 𝑦 to the power of negative three as one over 𝑦 cubed. Next, we can notice that negative three 𝐶 is simply another constant since both negative three and 𝐶 are constants. Therefore, we can rewrite it as 𝐷.
And this is a good point to use our boundary conditions in order to find the value of that constant 𝐷. We substitute in the values 𝑦 equals negative one and 𝑥 is equal to one-third. We have one over negative one cubed is equal to negative three multiplied by one-third plus 𝐷. Simplifying, we get negative one is equal to negative one plus 𝐷 giving us that 𝐷 is equal to zero. We can substitute this value of 𝐷 back into our equation to find that our specific solution is one over 𝑦 cubed is equal to negative three 𝑥.
Next, we can make 𝑦 the subject of this equation. We take the cube root of either side and then take the reciprocals of both sides to find that 𝑦 is equal to one over the cube root of negative three 𝑥. Our final step here is to substitute in the value 𝑥 is equal to nine. We can simplify this in order to obtain that 𝑦 is equal to one over the cube root of negative 27.
Since negative three cubed is equal to negative 27, we find that our solution is 𝑦 is equal to negative one-third.
