Solve \[xy\dfrac{{dy}}{{dx}} = {x^2} + 2{y^2}\]

Hint: This is the homogeneous differential equation and we have to solve this. First, we will substitute the values wherever required and then after differentiating the substituted value, we will integrate them. The formulas of logarithm and integration should be known to you to solve this question.

Complete step by step solution:
We have to solve the given expression
\[xy\dfrac{{dy}}{{dx}} = {x^2} + 2{y^2}\] ………eq(1)
This is the form of a homogeneous equation because the degree of the numerator is greater than the degree of the denominator.
First, we will take \[xy\] on the left-hand side of the equation and in the denominator of the right-hand side of the equation which is as shown below.
\[\dfrac{{dy}}{{dx}} = \dfrac{{{x^2} + 2{y^2}}}{{xy}},\]
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{{x^2}}}{{xy}} + \dfrac{{2{y^2}}}{{xy}}\],
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{x}{y} + \dfrac{{2y}}{x}\] ……..eq(2)
As this was a homogeneous equation, so we will solve it using the substitution method.
In the place of \[\dfrac{y}{x}\], we will put ‘ $u$ ’.
Let, \[\dfrac{y}{x} = u\]
\[ \Rightarrow y = xu\] ……..eq(3)
Now we will differentiate eq(2) with respect to $x$ which is as follows.
\[\dfrac{{dy}}{{dx}} = x\dfrac{{du}}{{dx}} + u\dfrac{{dx}}{{dx}}\] ,
\[ \Rightarrow \dfrac{{dy}}{{dx}} = x\dfrac{{du}}{{dx}} + u\] ,
Now we will put these values in eq(2) and the following results will be obtained.
\[\dfrac{{dy}}{{dx}} = \dfrac{x}{y} + \dfrac{{2y}}{x}\]
\[ \Rightarrow x\dfrac{{du}}{{dx}} + u = \dfrac{1}{u} + 2u\],
Now we will takethe like variables on one side of the equation and will further solve our question.
\[
  x\dfrac{{du}}{{dx}} = \dfrac{1}{u} + 2u - u, \\
   \Rightarrow x\dfrac{{du}}{{dx}} = \dfrac{1}{u} + u \\
    \\
 \]
Now we will take the LCM of the equation on the right-hand side.
\[x\dfrac{{du}}{{dx}} = \dfrac{{1 + {u^2}}}{u}\],
Now we will do the cross multiplication and then we will integrate the expression on both sides which are as shown below.
\[\int {\dfrac{{udu}}{{1 + {u^2}}} = \int {\dfrac{{dx}}{x}} } \] ……eq(4)
Integration of the expression \[\int {\dfrac{{udu}}{{1 + {u^2}}}} \] will be complicated, so we will do this integration by substituting the values.
Let, \[1 + {u^2} = t\]
Now we will do the differentiation of the above expression with respect to $x$ and we get the following results.
\[2u\dfrac{{du}}{{dx}} = \dfrac{{dt}}{{dx}},\]
\[ \Rightarrow 2udu = dt\],
\[ \Rightarrow udu = \dfrac{{dt}}{2}\]
We will put all these values in eq(4) and we get the below expression.
\[\int {\dfrac{{dt}}{{2t}} = \int {\dfrac{{dx}}{x}} } \] ….eq(5)
According to the formula of integration, we know that,
\[\int {\dfrac{{dx}}{x} = \log x} \]
On putting this formula in eq(5),
\[
  \int {\dfrac{{dt}}{{2t}} = \int {\dfrac{{dx}}{x}} } , \\
   \Rightarrow \dfrac{1}{2}\log t = \log x + \log c \\
 \] …………eq(6)
According to the formula of the logarithm,
\[\log {m^n} = n\log m\]
So eq(6) will become,
\[\log {t^{\dfrac{1}{2}}} = \log x + \log c\] …….eq(7)


And we also know that,
\[\log mn = \log m + \log n\]
On using this formula in eq(7), it becomes
\[\log {t^{\dfrac{1}{2}}} = \log xc\]
\[ \Rightarrow \log \sqrt t = \log xc\]
Logarithm will cancel out each other on both sides and the equation becomes.
\[\sqrt t = xc\] …….eq(8)


Now we will put the value of t in eq(8).
\[\sqrt {1 + {u^2}} = xc\]
And after that, we will put the value of ‘u’ in the above expression.
\[
  \sqrt {1 + \dfrac{{{y^2}}}{{{x^2}}}} = xc, \\
   \Rightarrow \sqrt {\dfrac{{{x^2} + {y^2}}}{{{x^2}}}} = xc, \\
 \]
\[ \Rightarrow \sqrt {{x^2} + {y^2}} = {x^2}c\]
On squaring both the sides of the above equation, we get
\[
  {x^2} + {y^2} = {x^4}{c^2} \\
    \\
 \]
Hence this is the solution.

Note: A homogeneous differential equation is of two types. The first-order differential equation will have the degree same in the numerator and denominator and in the second-order differential equation, the degree of the numerator will be greater than the degree of the denominator.