Question Video: Solving a Separable First Order Differential Equation given in the Normal Form and given Its Initial Value | Nagwa Question Video: Solving a Separable First Order Differential Equation given in the Normal Form and given Its Initial Value | Nagwa

# Question Video: Solving a Separable First Order Differential Equation given in the Normal Form and given Its Initial Value Mathematics • Higher Education

A relation π(π₯, π¦) = 0 is implicitly differentiated to obtain (dπ¦/dπ₯) = (2π₯ + 5)/(2π¦ + 5). Find the relation given that when π¦ = 3, π₯ = 3.

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### Video Transcript

A relation π of π₯, π¦ is equal to zero is implicitly differentiated to obtain dπ¦ by dπ₯ is equal to two π₯ plus five divided by two π¦ plus five. Find the relation given that when π¦ is equal to three, π₯ is equal to three.

We implicitly differentiate the relation π of π₯, π¦ is equal to zero to get that dπ¦ by dπ₯ is equal to two π₯ plus five divided by two π¦ plus five. We recall we call a differential equation separable if itβs a first-order differential equation which is the product of a function in π₯ with a function in π¦. And we can actually see the differential equation given to us in this case is a separable differential equation. We just set π of π₯ equal to two π₯ plus five and β of π¦ equal to one divided by two π¦ plus five.

So, we see that dπ¦ by dπ₯ is equal to π of π₯ times β of π¦. To solve a separable differential equation, we need to divide both sides of this equation by β of π¦. And since β of π¦ is a fraction, dividing by one divided by two π¦ plus five is the same as multiplying by two π¦ plus five. This gives us two π¦ plus five multiplied by dπ¦ by dπ₯ is equal to two π₯ plus five.

At this point, we remember dπ¦ by dπ₯ is definitely not a fraction. However, we can treat it a little bit like a fraction when weβre solving a separable differential equation. This gives us the equivalent statement two π¦ plus five dπ¦ is equal to two π₯ plus five dπ₯. And we can integrate both sides of this equation to get that the integral of two π¦ plus five with respect to π¦ is equal to the integral of two π₯ plus five with respect to π₯.

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 π¦ squared over two plus five π¦ plus a constant of integration π one is equal to two π₯ squared over two plus five π₯ plus the constant of integration we will call π two. We can simplify this. Two π¦ squared over two is equal to π¦ squared, and two π₯ squared over two is equal to π₯ squared. Then, we can combine the constants π one and π two into a new constant we will call π. This gives us the new equation π¦ squared plus five π¦ is equal to π₯ squared plus five π₯ plus π.

From the question, weβre told when π¦ is equal to three, π₯ is equal to three. So, letβs substitute π¦ is equal to three and π₯ is equal to three into this equation to find the value of π. Substituting π¦ is equal to three and π₯ is equal to three gives us three squared plus five times three is equal to three squared plus five times three plus π. And if we evaluate this expression and then simplify, we see that π is equal to zero.

So, using the fact that π is equal to zero, we have π¦ squared plus five π¦ is equal to π₯ squared plus five π₯. And since weβre told originally that we want a relation π of π₯, π¦ is equal to zero, we can subtract π¦ squared and five π¦ from both sides of this equation to get π₯ squared plus five π₯ minus π¦ squared minus five π¦ is equal to zero.

Therefore, weβve shown if a relation π of π₯, π¦ is equal to zero is implicitly differentiated to obtain dπ¦ by dπ₯ is equal to two π₯ plus five over two π¦ plus five. And when π¦ is equal to zero, π₯ is equal to three. Then we must have that π₯ squared plus five π₯ minus π¦ squared minus five π¦ is equal to zero.

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