Question

How can I solve this differential equation? : $x{{y}^{2}}\dfrac{dy}{dx}={{y}^{3}}-{{x}^{3}}$

Answer 1

Hint: We first need to divide both the sides of the given equation by $x{{h}^{2}}$ so that it will become $\dfrac{dy}{dx}=\dfrac{y}{x}-{{\left( \dfrac{x}{y} \right)}^{2}}$. Then, we have to substitute $\dfrac{y}{x}=v$ into this equation so that we will obtain a simpler equation \[x\dfrac{dv}{dx}=-\dfrac{1}{{{v}^{2}}}\] which can be solved easily by using the method of separation of variables. Finally, we have to back substitute the assumed variable $v=\dfrac{y}{x}$ and separate y in terms of x to get the final solution of the given differential equation.

Complete step by step solution:
The differential equation given in the above question is
$\Rightarrow x{{y}^{2}}\dfrac{dy}{dx}={{y}^{3}}-{{x}^{3}}$
Dividing both the sides of the above equation by $x{{y}^{2}}$, we get
$\begin{align}
  & \Rightarrow \dfrac{dy}{dx}=\dfrac{{{y}^{3}}-{{x}^{3}}}{x{{y}^{2}}} \\
 & \Rightarrow \dfrac{dy}{dx}=\dfrac{{{y}^{3}}}{x{{y}^{2}}}-\dfrac{{{x}^{3}}}{x{{y}^{2}}} \\
 & \Rightarrow \dfrac{dy}{dx}=\dfrac{y}{x}-\dfrac{{{x}^{2}}}{{{y}^{2}}} \\
 & \Rightarrow \dfrac{dy}{dx}=\dfrac{y}{x}-{{\left( \dfrac{x}{y} \right)}^{2}}........\left( i \right) \\
\end{align}$
Let us put $\dfrac{y}{x}$ to a variable $v$, so that we can write
\[\begin{align}
  & \Rightarrow \dfrac{y}{x}=v......\left( ii \right) \\
 & \Rightarrow y=xv \\
\end{align}\]
Differentiating both sides with respect to x, we get
$\Rightarrow \dfrac{dy}{dx}=v+x\dfrac{dv}{dx}......\left( iii \right)$
Substituting (ii) and (ii) in (i) we get
\[\begin{align}
  & \Rightarrow v+x\dfrac{dv}{dx}=v-{{\left( \dfrac{1}{v} \right)}^{2}} \\
 & \Rightarrow v+x\dfrac{dv}{dx}=v-\dfrac{1}{{{v}^{2}}} \\
\end{align}\]
Subtracting \[v\] from both sides of the above equation, we get
\[\begin{align}
  & \Rightarrow v+x\dfrac{dv}{dx}-v=v-\dfrac{1}{{{v}^{2}}}-v \\
 & \Rightarrow x\dfrac{dv}{dx}=-\dfrac{1}{{{v}^{2}}} \\
\end{align}\]
Multiplying both sides by \[{{v}^{2}}\], we get
\[\begin{align}
  & \Rightarrow {{v}^{2}}\left( x\dfrac{dv}{dx} \right)={{v}^{2}}\left( -\dfrac{1}{{{v}^{2}}} \right) \\
 & \Rightarrow x{{v}^{2}}\dfrac{dv}{dx}=-1 \\
\end{align}\]
Multiplying \[dx\] both sides
\[\Rightarrow x{{v}^{2}}dv=-dx\]
Dividing by \[x\] both sides
\[\begin{align}
  & \Rightarrow \dfrac{x{{v}^{2}}dv}{x}=-\dfrac{dx}{x} \\
 & \Rightarrow {{v}^{2}}dv=-\dfrac{dx}{x} \\
\end{align}\]
Integrating both the sides, we have
\[\begin{align}
  & \Rightarrow \int{{{v}^{2}}dv}=-\int{\dfrac{dx}{x}} \\
 & \Rightarrow \dfrac{{{v}^{3}}}{3}=-\ln x+C \\
\end{align}\]
Substituting $v=\dfrac{y}{x}$ we get
$\Rightarrow \dfrac{{{y}^{3}}}{3{{x}^{3}}}=-\ln x+C$
Finally, multiplying by $3{{x}^{3}}$ both sides, we get
\[\begin{align}
  & \Rightarrow \dfrac{{{y}^{3}}}{3{{x}^{3}}}\left( 3{{x}^{3}} \right)=\left( -\ln x+C \right)\left( 3{{x}^{3}} \right) \\
 & \Rightarrow {{y}^{3}}=-3{{x}^{3}}\ln x+3C{{x}^{3}} \\
 & \Rightarrow {{y}^{3}}=-3{{x}^{3}}\ln x+{{C}_{1}}{{x}^{3}} \\
\end{align}\]
(where \[{{C}_{1}}=3C\])
Hence, the given differential equation has been solved.

Note: In the above solution, we could substitute $\dfrac{y}{x}=v$ only since the given equation was homogenous, that is, we could express $\dfrac{dy}{dx}$ equal to some function of $\dfrac{y}{x}$. If an equation is non-homogenous, then we cannot use this method. After performing the integration on both sides, do not forget to back substitute $v=\dfrac{y}{x}$ into the obtained solution.