Question

The solution of a differential equation $\left( kx-{{y}^{2}} \right)dy=\left( {{x}^{2}}-ky \right)dx$ is:
(a) ${{x}^{3}}-{{y}^{3}}=3kyx+C$
(b) ${{x}^{3}}+{{y}^{3}}=3kyx+C$
(c) ${{x}^{2}}-{{y}^{2}}=2kyx+C$
(d) ${{x}^{2}}+{{y}^{2}}=2kyx+C$
(e) ${{x}^{3}}-{{y}^{2}}=3kyx+C$

Answer 1

Hint: Break the terms and adjust them in a manner such that they become an exact differential equation i.e in the form $I(x,y)dx+J(x,y)dy=0$.An implicit first-order ordinary differential equation of the form $I(x,y)dx+J(x,y)dy=0$, is called an exact differential equation.Then integrate the terms with respect to their respective variables to get the answer.

“Complete step-by-step answer:”
In mathematics, a differential equation is an equation that relates one or more functions and their derivatives. In applications, the function generally represents physical quantities, the derivatives represent their rates of change, and the differential equation defines a relationship between the two. Mainly the study of differential equations consists of the study of their solutions, the set of functions that satisfy each equation, and of the properties of their solutions. There are many types of differential equations but here we have to deal with ‘exact differential equations’.
An exact differential equation or total differential equation is a certain kind of ordinary differential equation which is widely used in physics and engineering. An implicit first-order ordinary differential equation of the form $I(x,y)dx+J(x,y)dy=0$, is called an exact differential equation.
Now, we come to the question. we have been given $\left( kx-{{y}^{2}} \right)dy=\left( {{x}^{2}}-ky \right)dx$. Now, separating the terms we get, $kxdy-{{y}^{2}}dy={{x}^{2}}dx-kydx$. This can be written as, $kxdy+kydx={{x}^{2}}dx+{{y}^{2}}dy$.
Now, $kxdy+kydx=k\left( xdy+ydx \right)=kd\left( xy \right)$
Therefore, $kd(xy)={{x}^{2}}dx+{{y}^{2}}dy$. Now integrating both sides we get,
$\begin{aligned} & \int{kd(xy)}=\int{{{x}^{2}}dx+\int{{{y}^{2}}dy}} \\\ & k\int{d(xy)}=\int{{{x}^{2}}dx+\int{{{y}^{2}}dy}} \\\ & kxy+c=\dfrac{{{x}^{3}}}{3}+\dfrac{{{y}^{3}}}{3} \\\ & 3kxy+C={{x}^{3}}+{{y}^{3}} \\\ & \Rightarrow {{x}^{3}}+{{y}^{3}}=3kxy+C \\\ \end{aligned}$
Hence, option (b) is the correct answer.

Note: First we have considered ‘$c$’ as the indefinite integral constant and later on ‘$C$’ because when we cross-multiplied ‘$3$’ in the solution, this ‘$c$’ will get converted into ‘$3c$’ which is again only a constant and is therefore represented by ‘$C$’. So, don’t get confuse.Students should know the important integration formulas while solving these types of problems.