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Question: Solve: \(\dfrac{{dy}}{{dx}} = \dfrac{1}{{{y^2} + \sin y}},y \ne 0\)...

Solve: dydx=1y2+sin⁡y,y≠0\dfrac{{dy}}{{dx}} = \dfrac{1}{{{y^2} + \sin y}},y \ne 0

Explanation

Solution

In the given question, we are given a differential equation. So, we have to solve the given differential equation using methods of integration. Indefinite integration gives us a family of curves. We have to find the family of curves that is represented by the given differential equation. So, we will first separate the variables and then do integration on both sides of the equation.

Complete step by step answer:
The given question requires us to solve a differential equation dydx=1y2+sin⁡y\dfrac{{dy}}{{dx}} = \dfrac{1}{{{y^2} + \sin y}} using methods of integration. First of all, we have to classify the type of differential equation.Now, we observe that we can separate the variables in the differential equation given to us. So, the differential equation dydx=1y2+sin⁡y\dfrac{{dy}}{{dx}} = \dfrac{1}{{{y^2} + \sin y}} is of variable separable form.Now, we separate the variables. So, we get,
⇒(y2+sin⁡y)dy=dx\Rightarrow \left( {{y^2} + \sin y} \right)dy = dx

Now, integrate both sides.
⇒∫(y2+sin⁡y)dy=∫dx\Rightarrow \int {\left( {{y^2} + \sin y} \right)dy} = \int {dx}
We know that the integral of sin⁡x\sin x with respect to x is −cos⁡x - \cos x. We also know the power rule of integration as ∫xndx=xn+1n+1\int {{x^n}dx} = \dfrac{{{x^{n + 1}}}}{{n + 1}}. So, we get,
⇒∫y2dy−cos⁡y=∫dx\Rightarrow \int {{y^2}dy} - \cos y = \int {dx}
Now, we also know the power rule of integration. We can write 11 as y0{y^0} and integrate using the power rule of integration ∫xndx=(xn+1n+1)\int {{x^n}dx} = \left( {\dfrac{{{x^{n + 1}}}}{{n + 1}}} \right). So, we get,
⇒y33−cos⁡y=x+c\Rightarrow \dfrac{{{y^3}}}{3} - \cos y = x + c where c is any arbitrary constant.

So, the general solution of the differential equation given to us dydx=1y2+sin⁡y,y≠0\dfrac{{dy}}{{dx}} = \dfrac{1}{{{y^2} + \sin y}},y \ne 0 is y33−cos⁡y=x+c\dfrac{{{y^3}}}{3} - \cos y = x + c.

Note: The indefinite integrals of certain functions may have more than one answer in different forms. However, all these forms are correct and interchangeable into one another. Indefinite integral gives us the family of curves as we don’t know the exact value of the constant. Exact value of the arbitrary constant can be calculated only if we are given a point lying on the function. We should remember the integrals of some basic trigonometric functions in order to solve the problem.