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

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

Find a relation between π’ and π‘ given that dπ’/dπ‘ = (1 + π‘β΄)/(π’π‘Β² + π’β΄ π‘Β²).

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### Video Transcript

Find a relation between π’ and π‘ given that dπ’ by dπ‘ equals one plus π‘ to the fourth power over π’π‘ squared plus π’ to the fourth power times π‘ squared.

Now, this differential equation might look really nasty. But it is, in fact, a separable differential equation. In this case, thatβs one where an expression for dπ’ by dπ‘ can be written as some function of π’ times some function of π‘. So how are we going to achieve that? Well, we begin by factoring the denominator by π‘ squared. And we find that dπ’ by dπ‘ is equal to one plus π‘ to the fourth power over π‘ squared times π’ plus π’ to the fourth power. We can now write this as some function of π‘ times some function of π’. Itβs one plus π‘ to the fourth power over π‘ squared times one over π’ plus π’ to the fourth power.

Letβs begin by multiplying both sides of this equation by π’ plus π’ to the fourth power. Then, we recall that whilst dπ’ by dπ‘ is not a fraction, we treat it a little like one. And we can say that π’ plus π’ to the fourth power dπ’ is equal to one plus π‘ to the fourth power over π‘ squared dπ‘. And thatβs great because weβre now ready to integrate both sides of our equation. The left-hand side is quite straightforward to integrate. The integral of π’ is π’ squared over two. And the integral of π’ to the fourth power is π’ to the fifth power over five. Donβt forget, since this is an indefinite integral, we need that constant of integration π.

On the right-hand side, weβre going to rewrite our integrand as one over π‘ squared plus π‘ to the fourth power over π‘ squared. And that simplifies to one over π‘ squared plus π‘ squared. But, of course, one over π‘ squared is the same as π‘ to the power of negative two. And we can now integrate as normal. When we integrate π‘ to the power of negative two, we add one to the exponent to get negative one and then we divide by that number. π‘ squared becomes π‘ cubed over three. And we have a second constant of integration; Iβve called that π. Now, of course, we can rewrite π‘ to the power of negative one over negative one as negative one over π‘.

Our final step is to subtract our constant π from both sides of the equation. That gives us a new constant π. Thatβs just the difference between π and π. So weβve found a relation between π’ and π‘. Itβs π’ to the fifth power over five plus π’ squared over two equals negative one over π‘ plus two cubed over three plus π.

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