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,

& \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{aligned}$$ $$\Rightarrow \dfrac{dM}{dy}=(-4x-4y)$$ …………………………..(6) Differentiating equation (4) with respect to x we get, $$\begin{aligned} & \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{aligned}$$ $$\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{aligned} & =\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{aligned}$$ $$\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{aligned} & \int{Ndy} \\\ & =\int{{{y}^{2}}dy} \\\ & =\dfrac{{{y}^{3}}}{3} \\\ \end{aligned}$$ $$\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{aligned} & \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{aligned}$$ Hence, the solution of the differential equation $$\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}$$ .