Question

Find the general solution of the equation \[\left[ 2x\sin \left( \dfrac{y}{x} \right)+2x\tan \left( \dfrac{y}{x} \right)-y\cos \left( \dfrac{y}{x} \right)-y{{\sec }^{2}}\left( \dfrac{y}{x} \right) \right]dx+\left[ x\cos \left( \dfrac{y}{x} \right)+x{{\sec }^{2}}\left( \dfrac{y}{x} \right) \right]dy=0\]

Answer 1

Hint: In this type of question we have to use the concept of exact differential equations. We know that the differential equation in the form \[Mdx+Ndy=0\] where \[M\] and \[N\] are function of \[x\] and \[y\] are said to be an exact differential equation if it satisfies \[\dfrac{\partial M}{\partial y}=\dfrac{\partial N}{\partial x}\]. As we are related with partial differentiation with respect to \[x\] and \[y\] when we partially differentiate with respect to \[x\] keep \[y\] as a constant and vice versa. Also we know that if a differential equation is an exact then its solution is given by, \[\int\limits_{\text{y constant}}{Mdx+\int{\left( \text{Terms of N free from x} \right)dx=c}}\]

Complete step by step solution:
Now, we have to find the general solution of the equation \[\left[ 2x\sin \left( \dfrac{y}{x} \right)+2x\tan \left( \dfrac{y}{x} \right)-y\cos \left( \dfrac{y}{x} \right)-y{{\sec }^{2}}\left( \dfrac{y}{x} \right) \right]dx+\left[ x\cos \left( \dfrac{y}{x} \right)+x{{\sec }^{2}}\left( \dfrac{y}{x} \right) \right]dy=0\]
Comparing the given differential equation with \[Mdx+Ndy=0\] we get,
\[\Rightarrow M=\left[ 2x\sin \left( \dfrac{y}{x} \right)+2x\tan \left( \dfrac{y}{x} \right)-y\cos \left( \dfrac{y}{x} \right)-y{{\sec }^{2}}\left( \dfrac{y}{x} \right) \right]\] and \[N=x\cos \left( \dfrac{y}{x} \right)+x{{\sec }^{2}}\left( \dfrac{y}{x} \right)\] First we will check whether the given function is exact or not. For this let us consider,
\[\Rightarrow M=\left[ 2x\sin \left( \dfrac{y}{x} \right)+2x\tan \left( \dfrac{y}{x} \right)-y\cos \left( \dfrac{y}{x} \right)-y{{\sec }^{2}}\left( \dfrac{y}{x} \right) \right]\]
Partially differentiating with respect to y we get,
\[\begin{align}
  & \Rightarrow \dfrac{\partial M}{\partial y}=2x\cos \left( \dfrac{y}{x} \right)\dfrac{1}{x}+2x{{\sec }^{2}}\left( \dfrac{y}{x} \right)\dfrac{1}{x}+y\sin \left( \dfrac{y}{x} \right)\dfrac{1}{x}-\cos \left( \dfrac{y}{x} \right)-{{\sec }^{2}}\left( \dfrac{y}{x} \right)-y2\sec \left( \dfrac{y}{x} \right)\sec \left( \dfrac{y}{x} \right)\tan \left( \dfrac{y}{x} \right)\dfrac{1}{x} \\
 & \Rightarrow \dfrac{\partial M}{\partial y}=2\cos \left( \dfrac{y}{x} \right)+2{{\sec }^{2}}\left( \dfrac{y}{x} \right)+\left( \dfrac{y}{x} \right)\sin \left( \dfrac{y}{x} \right)-\cos \left( \dfrac{y}{x} \right)-{{\sec }^{2}}\left( \dfrac{y}{x} \right)-2\left( \dfrac{y}{x} \right){{\sec }^{2}}\left( \dfrac{y}{x} \right)\tan \left( \dfrac{y}{x} \right) \\
 & \Rightarrow \dfrac{\partial M}{\partial y}=\cos \left( \dfrac{y}{x} \right)+{{\sec }^{2}}\left( \dfrac{y}{x} \right)+\left( \dfrac{y}{x} \right)\sin \left( \dfrac{y}{x} \right)-2\left( \dfrac{y}{x} \right){{\sec }^{2}}\left( \dfrac{y}{x} \right)\tan \left( \dfrac{y}{x} \right)\text{ }\cdots \cdots \cdots \text{e}{{\text{q}}^{\text{n}}}\left( 1 \right) \\
\end{align}\]
Now, let us consider,
\[\Rightarrow N=x\cos \left( \dfrac{y}{x} \right)+x{{\sec }^{2}}\left( \dfrac{y}{x} \right)\]
Partially differentiating with respect to x we get,
\[\begin{align}
  & \Rightarrow \dfrac{\partial N}{\partial x}=\cos \left( \dfrac{y}{x} \right)-x\sin \left( \dfrac{y}{x} \right)\left( \dfrac{-y}{{{x}^{2}}} \right)+{{\sec }^{2}}\left( \dfrac{y}{x} \right)+x2\sec \left( \dfrac{y}{x} \right)\sec \left( \dfrac{y}{x} \right)\tan \left( \dfrac{y}{x} \right)\left( \dfrac{-y}{{{x}^{2}}} \right) \\
 & \Rightarrow \dfrac{\partial N}{\partial x}=\cos \left( \dfrac{y}{x} \right)+\left( \dfrac{y}{x} \right)\sin \left( \dfrac{y}{x} \right)+{{\sec }^{2}}\left( \dfrac{y}{x} \right)-2\left( \dfrac{y}{x} \right){{\sec }^{2}}\left( \dfrac{y}{x} \right)\tan \left( \dfrac{y}{x} \right)\text{ }\cdots \cdots \cdots \text{e}{{\text{q}}^{\text{n}}}\left( 2 \right) \\
\end{align}\]
From \[\text{e}{{\text{q}}^{\text{n}}}\left( 1 \right)\] and \[\text{e}{{\text{q}}^{\text{n}}}\left( 2 \right)\] we can write,
\[\dfrac{\partial M}{\partial y}=\dfrac{\partial N}{\partial x}\]
Hence, the given differential equation is an exact differential equation.
Now, to find the solution of the differential equation consider,
\[\Rightarrow \int\limits_{\text{y constant}}{Mdx+\int{\left( \text{Terms of N free from x} \right)dx=c}}\]
We can observe that there is no any term in N which is free from x and hence we have to consider the first term only. Hence the formula changes as,
\[\Rightarrow \int\limits_{\text{y constant}}{\left[ 2x\sin \left( \dfrac{y}{x} \right)+2x\tan \left( \dfrac{y}{x} \right)-y\cos \left( \dfrac{y}{x} \right)-y{{\sec }^{2}}\left( \dfrac{y}{x} \right) \right]dx=c}\]
Now, we have to solve these integration, for this we use the relation between integration and derivative i.e. integration and derivative are inverse processes of each other.
As we can write the above term present in integration as \[\begin{align}
  & \dfrac{d}{dx}\left( {{x}^{2}} \right)\left[ \sin \left( \dfrac{y}{x} \right)+\tan \left( \dfrac{y}{x} \right) \right]=\left( {{x}^{2}} \right)\left[ \cos \left( \dfrac{y}{x} \right)\left( \dfrac{-y}{{{x}^{2}}} \right)+{{\sec }^{2}}\left( \dfrac{y}{x} \right)\left( \dfrac{-y}{{{x}^{2}}} \right) \right]+\left[ \sin \left( \dfrac{y}{x} \right)+\tan \left( \dfrac{y}{x} \right) \right]\left( 2x \right) \\
 & \dfrac{d}{dx}\left( {{x}^{2}} \right)\left[ \sin \left( \dfrac{y}{x} \right)+\tan \left( \dfrac{y}{x} \right) \right]=\left[ 2x\sin \left( \dfrac{y}{x} \right)+2x\tan \left( \dfrac{y}{x} \right)-y\cos \left( \dfrac{y}{x} \right)-y{{\sec }^{2}}\left( \dfrac{y}{x} \right) \right] \\
\end{align}\]
Hence, the above integration can be written as
\[\begin{align}
  & \Rightarrow \int\limits_{\text{y constant}}{\dfrac{d}{dx}\left( {{x}^{2}} \right)\left[ \sin \left( \dfrac{y}{x} \right)+\tan \left( \dfrac{y}{x} \right) \right]dx=c} \\
 & \Rightarrow \left( {{x}^{2}} \right)\left[ \sin \left( \dfrac{y}{x} \right)+\tan \left( \dfrac{y}{x} \right) \right]=c \\
\end{align}\]

Hence the general solution of the given differential equation is given by,
\[\left( {{x}^{2}} \right)\left[ \sin \left( \dfrac{y}{x} \right)+\tan \left( \dfrac{y}{x} \right) \right]=c\]

Note:
In this question students have to take care in finding the partial differentiation as well as in solving the integration. Also students have to take care during calculation as too many terms are present.