Question Video: Checking Which Is a Valid Solution of a Differential Equation on a Given Domain

Which of the following is a solution of π‘₯ + 𝑦𝑦′ = 0 defined for all βˆ’4 < π‘₯ < 4? [A] 𝑦 = √(4 βˆ’ π‘₯Β²) [B] 𝑦 = √(16 + π‘₯Β²) [C] 𝑦 = √(4 + π‘₯Β²) [D] 𝑦 = √(16 βˆ’ π‘₯Β²)

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Video Transcript

Which of the following is a solution of π‘₯ plus 𝑦 times 𝑦 prime is equal to zero defined for all π‘₯ greater than negative four and π‘₯ less than four? Option (A) 𝑦 is equal to the square root of four minus π‘₯ squared. Option (B) 𝑦 is equal to the square root of 16 plus π‘₯ squared. Option (C) 𝑦 is equal to the square root of four plus π‘₯ squared. Or option (D) 𝑦 is equal to the square root of 16 minus π‘₯ squared.

We’re given a differential equation and four possible solutions to this differential equation. We need to determine which of these four is a solution and which is defined for all values of π‘₯ on the open interval from negative four to four. Remember, we can check each of these individually to see if there are solutions to this differential equation. We want to substitute the expression for 𝑦 in each of our four options into our differential equation. We’ll see we need to find an expression for 𝑦 prime for each of these four options.

Let’s start with option (A). That’s 𝑦 is equal to the square root of four minus π‘₯ squared. The first thing we’ll do is rewrite this as four minus π‘₯ squared all raised to the power of one-half. We want to use this to find an expression for 𝑦 prime. That will be the derivative of 𝑦 with respect to π‘₯. And we have two ways of doing this. We could use the chain rule or we could use the general power rule. We’ll do this by using the general power rule.

We recall this tells us for a differentiable function 𝑓 of π‘₯ and constant 𝑛, the derivative of 𝑓 of π‘₯ all raised to the 𝑛th power with respect to π‘₯ is equal to 𝑛 times 𝑓 prime of π‘₯ multiplied by 𝑓 of π‘₯ raised to the power of 𝑛 minus one. In this case, our inner function 𝑓 of π‘₯ is four minus π‘₯ squared, which we know is differentiable because it’s a quadratic. And the value of our exponent 𝑛 is equal to one-half.

To apply the general power rule, we’re going to need to find an expression for the derivative of four minus π‘₯ squared with respect to π‘₯. We can do this by using the power rule for differentiation; we get negative two π‘₯. We’re now ready to apply the general power rule to find an expression for 𝑦 prime. It’s equal to one-half times negative two π‘₯ multiplied by four minus π‘₯ squared all raised to the power of one-half minus one. And we can simplify this. First, our exponent of one-half minus one can be simplified to give us negative one-half. Next, we have one-half multiplied by negative two is equal to negative one.

Finally, by using our laws of exponents, we can write four minus π‘₯ squared all raised to the power of negative one-half in our denominator using a square root symbol. This gives us for option (A) 𝑦 prime is equal to negative π‘₯ divided by the square root of four minus π‘₯ squared. We can then substitute our expression for 𝑦 and our expression for 𝑦 prime into our differential equation. Substituting these into our differential equation, we get that π‘₯ plus the square root of four minus π‘₯ squared multiplied by negative π‘₯ over the square root of four minus π‘₯ squared should be equal to zero.

And we can simplify this. We see in the second term we have a shared factor of the square root of four minus π‘₯ squared, so we can cancel these. So this simplifies to give us π‘₯ minus π‘₯ is equal to zero. And of course, this is true for all values of π‘₯. So what we’ve shown is 𝑦 is equal to the square root of four minus π‘₯ squared is a solution to our differential equation. However, remember, we want a solution which is defined for all values of π‘₯ in the open interval from negative four to four.

So what would happen if we were to substitute π‘₯ is equal to three into our solution to the differential equation? Doing this, we get that 𝑦 is equal to the square root of four minus three squared, which would simplify to give us that 𝑦 is equal to the square root of negative five. And since this doesn’t give us a real solution, this means that our solution (A) is not defined for all values of π‘₯ in the open interval from negative four to four. So option (A) can’t be the correct answer.

So we’ll clear some space and do exactly the same thing to check option (B). This time, we have 𝑦 prime will be the derivative of 16 plus π‘₯ squared all raised to the power of one-half with respect to π‘₯. And once again, we can differentiate this by using the general power rule. First, the derivative of our inner function is two π‘₯. This time, by applying the general power rule, we get that 𝑦 prime is equal to one-half times two π‘₯ multiplied by 16 plus π‘₯ squared all raised to the power of negative one-half.

And we can simplify this in the same way we did before. First, one-half multiplied by two is equal to one. Next, we’ll use our laws of exponents to rewrite 16 plus π‘₯ squared all raised to the power of negative one-half in our denominator with a square root symbol. And by doing this, we now have that 𝑦 prime is equal to π‘₯ divided by the square root of 16 plus π‘₯ squared. Once again, we can now substitute our expression for 𝑦 and 𝑦 prime into our differential equation. Substituting these into our differential equation, we get that π‘₯ plus the square root of 16 plus π‘₯ squared multiplied by π‘₯ divided by the square root of 16 plus π‘₯ squared should be equal to zero.

And we can simplify this. We have the square root of 16 plus π‘₯ squared divided by the square root of 16 plus π‘₯ squared is equal to one. And this simplifies to give us π‘₯ plus π‘₯ is equal to zero. And we can see this is not a solution to our differential equation. For example, when π‘₯ is equal to one, this would tell us that two should be equal to zero. So (B) is not the correct answer.

We could clear our working and then do the same for option (C). However, we notice the only difference for option (C) is the value of 16 should be four. However, if we were to just write four in every place in our working out where we had 16, we would see that none of our lines of working would change. The exact same process we did before will work for option (C). We can find an expression for 𝑦 prime by using the general power rule. We get 𝑦 prime is π‘₯ divided by the square root of four plus π‘₯ squared. And just as we did before, we can substitute our expressions for 𝑦 and 𝑦 prime into our differential equation. And we can simplify in exactly the same way to get π‘₯ plus π‘₯ is equal to zero. So option (C) can’t be the correct answer.

So by elimination, option (D) should be our correct answer. However, we’ll check that this is the case. So let’s clear some space and check that this is a solution to our differential equation. We’ll do this once again by using the general power rule. Differentiating our inner function of 16 minus π‘₯ squared, we get negative two π‘₯. Applying the general power rule, we get that 𝑦 prime is equal to one-half times negative two π‘₯ multiplied by 16 minus π‘₯ squared all raised to the power of negative one-half. Then, by simplifying this and using our laws of exponents, we get that 𝑦 prime is equal to negative π‘₯ divided by the square root of 16 minus π‘₯ squared.

Now, once again, we need to substitute our expressions for 𝑦 and 𝑦 prime into our differential equation. Doing this gives us that π‘₯ plus the square root of 16 minus π‘₯ squared multiplied by negative π‘₯ divided by the square root of 16 minus π‘₯ squared should be equal to zero. We’ll simplify this in exactly the same way we did before. We’ll cancel the shared factor of the square root of 16 minus π‘₯ squared. This gives us that π‘₯ minus π‘₯ should be equal to zero, which we know is true for any value of π‘₯. So we’ve shown that 𝑦 is equal to the square root of 16 minus π‘₯ squared is the solution to our differential equation. However, remember, we also need to check it’s defined for all values of π‘₯ on the open interval from negative four to four.

The first thing we note is we cancel the square roots of 16 minus π‘₯ squared with itself. This means the square root of 16 minus π‘₯ squared is not allowed to be equal to zero. And if this is not allowed to be equal to zero, then π‘₯ is not allowed to be equal to positive or negative four. And this is fine because we’re only interested in solutions valid on the open interval from negative four to four. Next, we’ll need to check the domain of our answer. For the square root of 16 minus π‘₯ squared to be defined, we must have that 16 minus π‘₯ squared is greater than or equal to zero. Of course, we’ve already dealt with the case where this is equal to zero. So we’ll just write this as 16 minus π‘₯ squared must be positive.

We can then just solve this inequality. We add π‘₯ squared to both sides and then either graphically by using what we know about square roots or by using what we know about the absolute value, we can see this is true if π‘₯ is greater than negative four and π‘₯ is less than four. This means that option (D) is in fact correct. And we were able to show that 𝑦 is equal to the square root of 16 minus π‘₯ squared is a solution to the differential equation π‘₯ plus 𝑦 times 𝑦 prime is equal to zero defined for all values of π‘₯ in the open interval from negative four to four.

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