Video: Solving Differential Equations

Is the function 𝑦 = 1/(2 + π‘₯) a solution to the differential equation 𝑦′ = βˆ’π‘¦Β²?

02:20

Video Transcript

Is the function 𝑦 equals one over two plus π‘₯ a solution to the differential equation 𝑦 prime equals negative 𝑦 squared?

Remember that 𝑦 prime is another way of saying d𝑦 by dπ‘₯, the first derivative of 𝑦 with respect to π‘₯. So, we’ve been given a first order differential equation, and we want to know whether the given function 𝑦 is a solution to it. That is, we need to know whether the function 𝑦 satisfies this equation.

Let’s begin then by first working out what 𝑦 prime, or d𝑦 by dπ‘₯, is equal to for this function 𝑦. And to do this, we can first express 𝑦 in an alternative form. We can write it as two plus π‘₯ to the power of negative one. We can then find this derivative using the general power rule, which says that if we have some function 𝑓 of π‘₯ to the power of 𝑛, then its derivative with respect to π‘₯ is equal to 𝑛 multiplied by 𝑓 prime of π‘₯ multiplied by 𝑓 of π‘₯ to the power of 𝑛 minus one.

Here, our function 𝑓 of π‘₯ is two plus π‘₯, and our power 𝑛 is negative one. So, applying the general power rule, we have 𝑛, that’s negative one, multiplied by the derivative of two plus π‘₯, which is just one, multiplied by 𝑓 of π‘₯. That’s two plus π‘₯, to the power of 𝑛 minus one. So, that’s the power of negative two. We can then rewrite this as negative one over two plus π‘₯ all squared. So, we know what the left-hand side of this differential equation would be for this function 𝑦.

On the right-hand side, we have negative 𝑦 squared. So, that’s the original function 𝑦 squared and then multiply it by negative one, which is equal to negative one over two plus π‘₯ all squared. To square a fraction, we can square the numerator and square the denominator. So, we have negative one squared, which is one, over two plus π‘₯ all squared.

Now, we compare our expressions for 𝑦 prime and negative 𝑦 squared. And we see that they are both equal to negative one over two plus π‘₯ all squared. And therefore, they are indeed equal to one another. This tells us that the function 𝑦 equals one over two plus π‘₯ does satisfy the given differential equation and, therefore, it is a solution.

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