Solve the Differential Equation y'=y^2 Sinx

In this tutorial we shall solve a differential equation of the form y' = {y^2}\sin x by using the separating the variables method.

The differential equation of the form is given as

y' = {y^2}\sin x

This differential equation can also be written as

\frac{{dy}}{{dx}} = {y^2}\sin x

Separating the variables, the given differential equation can be written as

\begin{gathered} \frac{1}{{{y^2}}}dy = \sin xdx \\ \Rightarrow {y^{ - 2}}dy = \sin xdx\,\,\,\,{\text{ - - - }}\left( {\text{i}} \right) \\ \end{gathered}

Keep in mind that in the separating variable technique the terms dy and dx are placed in the numerator with their respective variables.

Now integrating both sides of the equation (i), we have

\int {{y^{ - 2}}dy = \int {\sin xdx} }

Using the formulas of integration \int {{x^n}dx = \frac{{{x^{n + 1}}}}{{n + 1}}} and \int {\sin xdx = - \cos x} , we get

\begin{gathered} \frac{{{y^{ - 2 + 1}}}}{{ - 2 + 1}} = - \cos x + c \\ \Rightarrow \frac{{{y^{ - 1}}}}{{ - 1}} = - \cos x + c \\ \Rightarrow - \frac{1}{y} = - \cos x + c \\ \Rightarrow cy - y\cos x + 1 = 0 \\ \end{gathered}

This is the required solution of the given differential equation.