Question

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

Answer 1

Hint: First find the expression for $\dfrac{{dy}}{{dx}}$ in terms of $x$ and $y$. This will give us a homogeneous equation. Then substitute $y = vx$. Find the value of $\dfrac{{dy}}{{dx}}$ and put it in the above equation. Then solve the resulting differential equation using a variable separation method and put back the value of \[v\] to get the answer.

Complete step-by-step solution:
According to the question, we have to solve the differential equation \[\left( {{x^2} + xy} \right)dy = \left( {{x^2} + {y^2}} \right)dx\].
If we find the expression for $\dfrac{{dy}}{{dx}}$ in terms of $x$ and $y$ by cross multiplication, we’ll get:
$ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{{x^2} + {y^2}}}{{{x^2} + xy}}{\text{ }}.....{\text{(1)}}$
This is a homogeneous equation on the right hand side as the degree of all the terms in the equation is the same i.e. 2. So we’ll substitute $y = vx$ in the equation. We have:
$ \Rightarrow y = vx{\text{ }}.....{\text{(2)}}$
Differentiating it both sides, we’ll get:
$ \Rightarrow \dfrac{{dy}}{{dx}} = x\dfrac{{dv}}{{dx}} + v$
Putting these values in equation (1), we’ll get:
$
   \Rightarrow x\dfrac{{dv}}{{dx}} + v = \dfrac{{{x^2} + {{\left( {vx} \right)}^2}}}{{{x^2} + xvx}} \\
   \Rightarrow x\dfrac{{dv}}{{dx}} + v = \dfrac{{{x^2} + {v^2}{x^2}}}{{{x^2} + v{x^2}}} \\
   \Rightarrow x\dfrac{{dv}}{{dx}} + v = \dfrac{{1 + {v^2}}}{{1 + v}}
 $
On further simplification, this will give us:
$
   \Rightarrow x\dfrac{{dv}}{{dx}} = \dfrac{{1 + {v^2}}}{{1 + v}} - v \\
   \Rightarrow x\dfrac{{dv}}{{dx}} = \dfrac{{1 + {v^2} - v - {v^2}}}{{1 + v}} \\
   \Rightarrow x\dfrac{{dv}}{{dx}} = \dfrac{{1 - v}}{{1 + v}}
 $
Now we will cross multiply for variable separation, we’ll get:
$ \Rightarrow \dfrac{{1 + v}}{{1 - v}}dv = \dfrac{{dx}}{x}$
This can also be written as:
$ \Rightarrow \dfrac{{v + 1}}{{v - 1}}dv = - \dfrac{{dx}}{x}$
Integrating both sides, we’ll get:
$ \Rightarrow \int {\dfrac{{v + 1}}{{v - 1}}dv} = - \int {\dfrac{{dx}}{x}} $
Left hand side can be simplified as:
$
   \Rightarrow \int {\dfrac{{v - 1 + 2}}{{v - 1}}dv} = - \int {\dfrac{{dx}}{x}} \\
   \Rightarrow \int {\dfrac{{v - 1}}{{v - 1}}dv + 2\int {\dfrac{{dv}}{{v - 1}}} } = - \int {\dfrac{{dx}}{x}} \\
   \Rightarrow \int {dv + 2\int {\dfrac{{dv}}{{v - 1}}} } = - \int {\dfrac{{dx}}{x}}
 $
We know that the integration of $\dfrac{1}{x}$ with respect to $x$ is $\ln \left| x \right|$.
$ \Rightarrow \int {\dfrac{{dx}}{x}} = \ln \left| x \right|$
Similarly the integration of $\dfrac{1}{{v - 1}}$ with respect to $v$ is $\ln \left| {v - 1} \right|$.
$ \Rightarrow \int {\dfrac{{dv}}{{v - 1}}} = \ln \left| {v - 1} \right|$
Using these results in our integration, we’ll get:
$ \Rightarrow v + 2\ln \left| {v - 1} \right| = - \ln \left| x \right| + \ln C$, where $\ln C$ is the integration constant.
We have taken a logarithmic constant just to adjust our equation as we are getting other terms in logarithms also.
$ \Rightarrow v + 2\ln \left| {v - 1} \right| = \ln C - \ln \left| x \right|$
Now, we will use few properties of logarithm as shown below:
$
   \Rightarrow a\ln b = \ln {b^a} \\
   \Rightarrow \ln m - \ln n = \ln \dfrac{m}{n}
 $
Using these properties in our equation, we’ll get:
$
   \Rightarrow v + \ln {\left| {v - 1} \right|^2} = \ln \dfrac{C}{x} \\
   \Rightarrow \ln {\left( {v - 1} \right)^2} - \ln \dfrac{C}{x} = - v
 $
Again using property of logarithm, we will get:
$
   \Rightarrow \ln \left( {\dfrac{{{{\left( {v - 1} \right)}^2}}}{{\dfrac{C}{x}}}} \right) = - v \\
   \Rightarrow \ln \left( {\dfrac{{x{{\left( {v - 1} \right)}^2}}}{C}} \right) = - v
 $
Putting back $v = \dfrac{y}{x}$ from equation (2), we’ll get:
$
   \Rightarrow \ln \left( {\dfrac{{x{{\left( {\dfrac{y}{x} - 1} \right)}^2}}}{C}} \right) = - \dfrac{y}{x} \\
   \Rightarrow \ln \left( {\dfrac{{x\dfrac{{{{\left( {y - x} \right)}^2}}}{{{x^2}}}}}{C}} \right) = - \dfrac{y}{x} \\
   \Rightarrow \ln \dfrac{{{{\left( {y - x} \right)}^2}}}{{Cx}} = - \dfrac{y}{x}
 $
Taking anti logarithm both sides, we’ll get:
$
   \Rightarrow \dfrac{{{{\left( {y - x} \right)}^2}}}{{Cx}} = {e^{ - \dfrac{y}{x}}} \\
   \Rightarrow {\left( {y - x} \right)^2} = Cx{e^{ - \dfrac{y}{x}}}
 $

${\left( {y - x} \right)^2} = Cx{e^{ - \dfrac{y}{x}}}$ is the final solution of the differential equation.

Note: When the degree of all the terms in a polynomial equation is the same then those equations are called homogeneous equations. And if we have to solve the differential equation containing homogeneous equations then we substitute $y = vx$ and proceed as we did above.
If we can separate the two variables of the differential equation on either side of the equation then we can solve it using the variable separation method also. This we have done in the step just before integration in the above problem separating $v$ and $x$ on either side of the equation.