### Video Transcript

Is the function π¦ equals π₯ to the fourth power over four a solution to the differential equation π¦ prime equals π₯ cubed?

Recall that a differential equation is an equation which relates some function with its derivatives. The differential equation in this question is π¦ prime equals π₯ cubed. Recall that π¦ prime is just another way of saying dπ¦ by dπ₯, the first derivative of π¦ with respect to π₯. And weβve been asked to check whether this function, π¦ equals π₯ to the fourth power over four, is a solution to this differential equation.

To do this, weβre going to need to make a substitution using this function into a differential equation. A possible solution is π¦ equals π₯ to the fourth power over four, which we could equivalently write as one over four multiplied by π₯ to the fourth power. And we can differentiate this to find π¦ prime. We firstly recall the power rule, which tells us that the derivative with respect to π₯ of π₯ to the πth power is equal to π multiplied by π₯ to the power of π minus one. And using this, we find that π¦ prime equals one over four multiplied by four π₯ cubed. But we can actually simplify this to π₯ cubed.

So, using the possible solution, we find that the left-hand side is π₯ cubed. And we see that the left-hand side of the differential equation and the right-hand side of the differential equation agree. Therefore, we can conclude that this is a solution to our differential equation.