Question

How to solve the homogeneous differential equation (2xy+3y^2)dx-(2xy+x^2)dy=0

Answer 1

Answer:

this is correct..............................

Answer 2

Given,

$\[\left( {2xy + 3{y^2}} \right)dx - \left( {2xy + {x^2}} \right)dy = 0\]$

Solution,

Rewrite the given homogeneous differential equation.

$\[ \Rightarrow \frac{{dy}}{{dx}} = \frac{{2xy + 3{y^2}}}{{2xy + {x^2}}} - - - - \left( 1 \right)\]$

$\[Put,y = vx\]$

$\[\frac{{dy}}{{dx}} = v + x\frac{{dv}}{{dx}}\]$

Put it in equation 1.

$\[ \Rightarrow v + x\frac{{dv}}{{dx}} = \frac{{2x\left( {vx} \right) + 3{{\left( {vx} \right)}^2}}}{{2x\left( {vx} \right) + {x^2}}}\]$

$\[ \Rightarrow x\frac{{dv}}{{dx}} = \frac{{2v{x^2} + 3{v^2}{x^2}}}{{2v{x^2} + {x^2}}} - v\]$

Solve to separate the variables.

$\[ \Rightarrow x\frac{{dv}}{{dx}} = \frac{{2v + 3{v^2}}}{{2v + 1}} - v\]$

$\[ \Rightarrow x\frac{{dv}}{{dx}} = \frac{{2v + 3{v^2} - 2{v^2} - v}}{{2v + 1}}\]$

$\[ \Rightarrow x\frac{{dv}}{{dx}} = \frac{{{v^2} + v}}{{2v + 1}}\]$

$\[ \Rightarrow \frac{{\left( {2v + 1} \right)dv}}{{\left( {{v^2} + v} \right)}} = \frac{{dx}}{x} - - - - \left( 2 \right)\]$

$\[\begin{array}{l}Take,t = {v^2} + v\\dt = \left( {2v + 1} \right)dv\end{array}\]$

Put in equation 2 and integrate.

$\[ \Rightarrow \int {\frac{{dt}}{t}} = \int {\frac{{dx}}{x}} \]$

$\[ \Rightarrow \ln t = \ln x + \ln c\]$

$\[ \Rightarrow t = xc\]$

Put value of t.

$\[ \Rightarrow {v^2} + v = xc\]$

Put value of v.

$\[ \Rightarrow {\left( {\frac{y}{x}} \right)^2} + \left( {\frac{y}{x}} \right) = xc\]$

$\[ \Rightarrow {y^2} + xy - {x^3}c = 0\]$, where c is an arbitrary constant.

Hence the solution of homogeneous differential equation is $\[{y^2} + xy - {x^3}c = 0\]$