Video Transcript
A relation 𝑓 of 𝑥, 𝑦 is equal to zero is implicitly differentiated to obtain d𝑦 by d𝑥 is equal to two 𝑥 plus five divided by two 𝑦 plus five. Find the relation given that when 𝑦 is equal to three, 𝑥 is equal to three.
We implicitly differentiate the relation 𝑓 of 𝑥, 𝑦 is equal to zero to get that d𝑦 by d𝑥 is equal to two 𝑥 plus five divided by two 𝑦 plus five. We recall we call a differential equation separable if it’s a first-order differential equation which is the product of a function in 𝑥 with a function in 𝑦. And we can actually see the differential equation given to us in this case is a separable differential equation. We just set 𝑔 of 𝑥 equal to two 𝑥 plus five and ℎ of 𝑦 equal to one divided by two 𝑦 plus five.
So, we see that d𝑦 by d𝑥 is equal to 𝑔 of 𝑥 times ℎ of 𝑦. To solve a separable differential equation, we need to divide both sides of this equation by ℎ of 𝑦. And since ℎ of 𝑦 is a fraction, dividing by one divided by two 𝑦 plus five is the same as multiplying by two 𝑦 plus five. This gives us two 𝑦 plus five multiplied by d𝑦 by d𝑥 is equal to two 𝑥 plus five.
At this point, we remember d𝑦 by d𝑥 is definitely not a fraction. However, we can treat it a little bit like a fraction when we’re solving a separable differential equation. This gives us the equivalent statement two 𝑦 plus five d𝑦 is equal to two 𝑥 plus five d𝑥. And we can integrate both sides of this equation to get that the integral of two 𝑦 plus five with respect to 𝑦 is equal to the integral of two 𝑥 plus five with respect to 𝑥.
We can evaluate both of these integrals by using the power rule for integration. We add one to the exponent and then divide by this new exponent. This gives us two 𝑦 squared over two plus five 𝑦 plus a constant of integration 𝑐 one is equal to two 𝑥 squared over two plus five 𝑥 plus the constant of integration we will call 𝑐 two. We can simplify this. Two 𝑦 squared over two is equal to 𝑦 squared, and two 𝑥 squared over two is equal to 𝑥 squared. Then, we can combine the constants 𝑐 one and 𝑐 two into a new constant we will call 𝑐. This gives us the new equation 𝑦 squared plus five 𝑦 is equal to 𝑥 squared plus five 𝑥 plus 𝑐.
From the question, we’re told when 𝑦 is equal to three, 𝑥 is equal to three. So, let’s substitute 𝑦 is equal to three and 𝑥 is equal to three into this equation to find the value of 𝑐. Substituting 𝑦 is equal to three and 𝑥 is equal to three gives us three squared plus five times three is equal to three squared plus five times three plus 𝑐. And if we evaluate this expression and then simplify, we see that 𝑐 is equal to zero.
So, using the fact that 𝑐 is equal to zero, we have 𝑦 squared plus five 𝑦 is equal to 𝑥 squared plus five 𝑥. And since we’re told originally that we want a relation 𝑓 of 𝑥, 𝑦 is equal to zero, we can subtract 𝑦 squared and five 𝑦 from both sides of this equation to get 𝑥 squared plus five 𝑥 minus 𝑦 squared minus five 𝑦 is equal to zero.
Therefore, we’ve shown if a relation 𝑓 of 𝑥, 𝑦 is equal to zero is implicitly differentiated to obtain d𝑦 by d𝑥 is equal to two 𝑥 plus five over two 𝑦 plus five. And when 𝑦 is equal to zero, 𝑥 is equal to three. Then we must have that 𝑥 squared plus five 𝑥 minus 𝑦 squared minus five 𝑦 is equal to zero.
