Question

Solve the differential equation, \[\left( {{x}^{2}}-4xy-2{{y}^{2}} \right)dx+\left( {{y}^{2}}-4xy-2{{x}^{2}} \right)dy=0\] .

Answer 1

Hint: The given differential equation is of the form \[Mdx+Ndy=0\] . On comparing we get \[M=\left( {{x}^{2}}-4xy-2{{y}^{2}} \right)\] and \[N=\left( {{y}^{2}}-4xy-2{{x}^{2}} \right)\] . We know that an equation is said to be an exact differential equation if it satisfies the condition \[\dfrac{dM}{dy}=\dfrac{dN}{dx}\] . The solution of the exact differential equation is given by \[\int{Mdx+\int{Ndy}=\text{Constant}}\] . Here, in \[\int{Mdx}\] , we integrate M and take the terms of y as a constant. In \[\int{Ndy}\] , we integrate those terms of N which do not contain any terms of x.

Complete step-by-step answer:
According to the question, our given differential equation is
\[\left( {{x}^{2}}-4xy-2{{y}^{2}} \right)dx+\left( {{y}^{2}}-4xy-2{{x}^{2}} \right)dy=0\] …………………………………..(1)
The given equation is of the form, \[Mdx+Ndy=0\] ……………………….(2)
Comparing equation (1) and equation (2), we get
\[M=\left( {{x}^{2}}-4xy-2{{y}^{2}} \right)\] ……………………….(3)
\[N=\left( {{y}^{2}}-4xy-2{{x}^{2}} \right)\] …………………………(4)
Check the given differential equation if it is of exact form or not.
The given equation is said to be an exact differential equation if it satisfies the condition,
\[\dfrac{dM}{dy}=\dfrac{dN}{dx}\] ……………………..(5)
Differentiating equation (3) with respect to y we get,
\[\begin{align}
  & \dfrac{dM}{dy}=\dfrac{d\left( {{x}^{2}}-4xy-2{{y}^{2}} \right)}{dy} \\
 & \Rightarrow \dfrac{dM}{dy}=\dfrac{d{{x}^{2}}}{dy}-\dfrac{d(4xy)}{dy}-\dfrac{d(2{{y}^{2}})}{dy} \\
 & \Rightarrow \dfrac{dM}{dy}=0-4x-4y \\
\end{align}\]
\[\Rightarrow \dfrac{dM}{dy}=(-4x-4y)\] …………………………..(6)
Differentiating equation (4) with respect to x we get,
\[\begin{align}
  & \dfrac{dN}{dx}=\dfrac{d\left( {{y}^{2}}-4xy-2{{x}^{2}} \right)}{dx} \\
 & \Rightarrow \dfrac{dN}{dx}=\dfrac{d{{y}^{2}}}{dx}-\dfrac{d(4xy)}{dx}-\dfrac{d(2{{x}^{2}})}{dx} \\
 & \Rightarrow \dfrac{dN}{dx}=0-4y-4x \\
\end{align}\]
\[\Rightarrow \dfrac{dN}{dy}=(-4x-4y)\] …………………………..(6)
From equation (5) and equation (6) we have the values of \[\dfrac{dM}{dx}\] and \[\dfrac{dN}{dy}\] .
\[\dfrac{dM}{dx}=\dfrac{dN}{dy}=(-4x-4y)\]
The given differential equation is an exact differential equation because it satisfies the condition,
\[\dfrac{dM}{dy}=\dfrac{dN}{dx}\] .
We know the steps to solve the exact form of differential equation. The steps to solve exact differential equation are,
Step 1st: We have to integrate M with respect to x keeping y as constant.
\[\int{Mdx}\]
From equation (3), we have M.
Now, integrating M with respect to x and taking y as constant.
\[\int{Mdx}\]
\[\begin{align}
  & =\int{\left( {{x}^{2}}-4xy-2{{y}^{2}} \right)dx} \\
 & =\int{{{x}^{2}}dx-4y\int{xdx-2{{y}^{2}}\int{dx}}} \\
 & =\dfrac{{{x}^{3}}}{3}-\dfrac{4y{{x}^{2}}}{2}-2{{y}^{2}}x \\
\end{align}\]
\[\int{Mdx}=\dfrac{{{x}^{3}}}{3}-2{{x}^{2}}y-2x{{y}^{2}}\] ……………………….(7)
Step 2nd: We have to integrate those terms of N which do contain x with respect to y.
From equation (4), we have N.
Now, integrating those terms of N which do contain x with respect to y, we get
\[\int{Ndy}\]
\[\begin{align}
  & \int{Ndy} \\
 & =\int{{{y}^{2}}dy} \\
 & =\dfrac{{{y}^{3}}}{3} \\
\end{align}\]
\[\int{Ndy}=\dfrac{{{y}^{3}}}{3}\] ………………….(8)
Step 3rd: We know the solution of the exact differential equation of the form \[Mdx+Ndy=0\] is
\[\int{Mdx+\int{Ndy}=\text{Constant}}\] ………………(9)
From equation (7), equation (8), and equation (9), we have
\[\begin{align}
  & \dfrac{{{x}^{3}}}{3}-2{{x}^{2}}y-2x{{y}^{2}}+\dfrac{{{y}^{3}}}{3}=\text{Constant} \\
 & \Rightarrow {{x}^{3}}+{{y}^{3}}-6{{x}^{2}}y-6x{{y}^{2}}=\text{Constant} \\
 & \Rightarrow {{x}^{3}}+{{y}^{3}}-6xy(x+y)=\text{Constant} \\
\end{align}\]
Hence, the solution of the differential equation \[\left( {{x}^{2}}-4xy-2{{y}^{2}} \right)dx+\left( {{y}^{2}}-4xy-2{{x}^{2}} \right)dy=0\] is \[{{x}^{3}}+{{y}^{3}}-6xy(x+y)=\text{Constant}\] .

Note: In this question, one may think that an equation is said to be an exact differential equation if it satisfies the condition, \[\dfrac{dM}{dx}=\dfrac{dN}{dy}\] which is wrong. The correct condition which an exact differential equation satisfies is \[\dfrac{dM}{dy}=\dfrac{dN}{dx}\] .