Question

Solve the following equation, ${{x}^{2}}ydx-\left( {{x}^{3}}+{{y}^{3}} \right)dy=0$.

Answer 1

Hint: In order to solve this question, we will put $x=vy$ after writing the given equation in $\dfrac{dy}{dx}$ form and we will simplify it to get an integrable form and then we will integrate it to get the answer. Also, we need to remember a few standard integrations like, $\int{{{x}^{n}}}dx=\dfrac{{{x}^{n+1}}}{n+1}$ and $\int{\dfrac{1}{x}}dx=\ln x+c$ . By using them, we can get our answer.

Complete step-by-step solution -
In this question, we have been given a differential equation, ${{x}^{2}}ydx-\left( {{x}^{3}}+{{y}^{3}} \right)dy=0$. To solve this equation, we have to represent it in $\dfrac{dy}{dx}$ form. For that, we will write the equation as,
$\begin{align}
  & {{x}^{2}}ydx-\left( {{x}^{3}}+{{y}^{3}} \right)dy=0 \\
 & \Rightarrow {{x}^{2}}ydx=\left( {{x}^{3}}+{{y}^{3}} \right)dy \\
\end{align}$
And further, we can write it as,
$\dfrac{dx}{dy}=\dfrac{{{x}^{3}}+{{y}^{3}}}{{{x}^{2}}y}$
Here, we can see that the right side of the equation is homogeneous. So, we will put $x=vy$, to simplify. And if we differentiate $x=vy$ with respect to y, we can use the product rule of differentiation, which is given by $\dfrac{d}{dy}\left( u.v \right)=u\dfrac{dv}{dy}+v\dfrac{du}{dy}$. Here, we have u = v and v = y, so we get,
$\dfrac{dx}{dy}=v\dfrac{dy}{dy}+y\dfrac{dv}{dy}$
So, we can say,
 $v+y\dfrac{dv}{dy}=\dfrac{{{\left( vy \right)}^{3}}+{{y}^{3}}}{{{\left( vy \right)}^{2}}y}$
And we can further write it as,
$v+y\dfrac{dv}{dy}=\dfrac{{{v}^{3}}{{y}^{3}}+{{y}^{3}}}{{{v}^{2}}{{y}^{3}}}$
And it is the same as,
$v+y\dfrac{dv}{dy}=\dfrac{\left( {{v}^{3}}+1 \right){{y}^{3}}}{{{v}^{2}}{{y}^{3}}}$
And we know that the common terms from the numerator and the denominator get cancelled. So, we get,
$v+y\dfrac{dv}{dy}=\dfrac{{{v}^{3}}+1}{{{v}^{2}}}$
Now, we will try to take all the terms of v on one side and the terms of y on the other side. So, we can write,
$\begin{align}
  & y\dfrac{dv}{dy}=\dfrac{{{v}^{3}}+1}{{{v}^{2}}}-v \\
 & \Rightarrow y\dfrac{dv}{dy}=\dfrac{{{v}^{3}}+1-{{v}^{3}}}{{{v}^{2}}} \\
 & \Rightarrow y\dfrac{dv}{dy}=\dfrac{1}{{{v}^{2}}} \\
 & \Rightarrow {{v}^{2}}dv=\dfrac{dy}{y} \\
\end{align}$
Now, we will integrate both the sides of the equation. So, we get,
$\int{{{v}^{2}}dv}=\int{\dfrac{dy}{y}}$
Now, we know that $\int{{{x}^{n}}}dx=\dfrac{{{x}^{n+1}}}{n+1}$. So, we can write $\int{{{v}^{2}}dv}=\dfrac{{{v}^{3}}}{3}$ and we know that $\int{\dfrac{dy}{y}}=\ln y+c$. So, we get the equation as,
$\dfrac{{{v}^{3}}}{3}=\ln y+c$
Now, we will put $v=\dfrac{x}{y}$, because we assumed that $x=vy$. So, we get,
$\dfrac{{{x}^{3}}}{3{{y}^{3}}}=\ln y+c$
Hence, we get the solution of ${{x}^{2}}ydx-\left( {{x}^{3}}+{{y}^{3}} \right)dy=0$ as $\dfrac{{{x}^{3}}}{3{{y}^{3}}}=\ln y+c$.

Note: While solving this question, we need to remember that after writing $\dfrac{dx}{dy}=\dfrac{{{x}^{3}}+{{y}^{3}}}{{{x}^{2}}y}$, we get the right side as the homogenous part and for solving a homogeneous differential equation, we put $x=vy$ and then we solve it. If we don’t use this method, then we may not be able to solve this question.