Question

What is the general solution of the differential equation
$\dfrac{{dy}}{{dx}} = \dfrac{{x + 2y - 3}}{{2x + y - 3}}?$

Answer 1

We perform a transformation to remove the constants from the numerator and denominator. First, we will consider the simultaneous equations as $x + 2y - 3 = 0$ and $2x + y - 3 = 0$ and we will get the values of $x$ and $y$ as $1$ . After that we will let $u = x - 1$ and $v = y - 1$ and find $\dfrac{{du}}{{dx}}$ and $\dfrac{{dv}}{{dy}}$ Then we will substitute the values in the given equation and apply the chain rule and hence we will get a transformed equation in $u$ and $v$ .After that we will use the substitution method and substitute $v = wu$ and find $\dfrac{{dv}}{{du}}$ and then substitute all the values in the transformed differential equation. After that we will apply integration and solve it using separating the variable method. And after simplification we will get the required result.

Complete step by step answer:
We have given:
$\dfrac{{dy}}{{dx}} = \dfrac{{x + 2y - 3}}{{2x + y - 3}}{\text{ }} - - - \left( A \right)$
First, we will perform a transformation to remove the constants from the numerator and denominator.
So, let us consider the simultaneous equations as:
$x + 2y - 3 = 0{\text{ }} - - - \left( 1 \right)$
$2x + y - 3 = 0{\text{ }} - - - \left( 2 \right)$
Now we will solve equation $\left( 1 \right)$ and $\left( 2 \right)$ using elimination method
So, multiply the equation $\left( 1 \right)$ by $2$
Therefore, we get
$2x + 4y - 6 = 0{\text{ }} - - - \left( 3 \right)$
Now, subtract equation $\left( 2 \right)$ from $\left( 3 \right)$ , we get
$3y - 3 = 0$
$\Rightarrow y = 1$
Now put the value of $y$ in equation $\left( 1 \right)$ we get
$x + 2 - 3 = 0$
$\Rightarrow x = 1$
Now as a result we will perform two linear transformations as:
Let
$u = x - 1{\text{ }} - - - \left( 4 \right)$
$v = y - 1{\text{ }} - - - \left( 5 \right)$
So, from equation $\left( 4 \right)$
$\dfrac{{du}}{{dx}} = 1$ and $x = u + 1{\text{ }} - - - \left( 6 \right)$
And from equation $\left( 5 \right)$
$\dfrac{{dv}}{{dy}} = 1$ and $y = v + 1{\text{ }} - - - \left( 7 \right)$
Now, substitute the values of $x$ and $y$ from equation $\left( 6 \right)$ and $\left( 7 \right)$ in equation $\left( A \right)$
Therefore, we get
$\dfrac{{dy}}{{dx}} = \dfrac{{\left( {u + 1} \right) + 2\left( {v + 1} \right) - 3}}{{2\left( {u + 1} \right) + \left( {v + 1} \right) - 3}}{\text{ }}$
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{u + 1 + 2v + 2 - 3}}{{2u + 2 + v + 1 - 3}}{\text{ }}$
After simplification, we get
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{u + 2v}}{{2u + v}}{\text{ }}$
Now using the chain rule, we can write
$\dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{dv}} \times \dfrac{{dv}}{{du}} \times \dfrac{{du}}{{dx}}$
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{dv}}{{du}}$
Thus, we get a transformed equation as,
$\Rightarrow \dfrac{{dv}}{{du}} = \dfrac{{u + 2v}}{{2u + v}}{\text{ }} - - - \left( B \right)$
Now we will use substitution method
So, substitute $v = wu$
On differentiating w.r.t $u$ using product rule, we get
$\dfrac{{dv}}{{du}} = \left( w \right)\left( {\dfrac{d}{{du}}u} \right) + \left( {\dfrac{d}{{du}}w} \right)\left( u \right)$
$\Rightarrow \dfrac{{dv}}{{du}} = w + u\dfrac{{dw}}{{du}}$
Using this substitution in equation $\left( B \right)$ we get
$w + u\dfrac{{dw}}{{du}} = \dfrac{{u + 2wu}}{{2u + wu}}$
$\Rightarrow u\dfrac{{dw}}{{du}} = \dfrac{{u + 2wu}}{{2u + wu}} - w$
On taking L.C.M we get
$\Rightarrow u\dfrac{{dw}}{{du}} = \dfrac{{\left( {u + 2wu} \right) - w\left( {2u + wu} \right)}}{{2u + wu}}$
$\Rightarrow u\dfrac{{dw}}{{du}} = \dfrac{{u + 2wu - 2uw - {w^2}u}}{{2u + wu}}$
On cancelling $2wu$ , we get
$\Rightarrow u\dfrac{{dw}}{{du}} = \dfrac{{u - {w^2}u}}{{2u + wu}}$
Taking common $u$ from both numerators and denominators we get
$\Rightarrow u\dfrac{{dw}}{{du}} = \dfrac{{u\left( {1 - {w^2}} \right)}}{{u\left( {2 + w} \right)}}$
$\Rightarrow u\dfrac{{dw}}{{du}} = \dfrac{{1 - {w^2}}}{{2 + w}}$
This is now a separable differential equation, so we can rearrange and separate the variable.
Therefore, we get
$\Rightarrow \dfrac{{2 + w}}{{1 - {w^2}}}dw = \dfrac{1}{u}du$
On integration, we get
$\Rightarrow \int {\dfrac{{2 + w}}{{1 - {w^2}}}dw} = \int {\dfrac{1}{u}du}$
$\Rightarrow \int {\dfrac{2}{{1 - {w^2}}}dw} + \int {\dfrac{w}{{1 - {w^2}}}dw} = \int {\dfrac{1}{u}du}$
$\Rightarrow \int {\dfrac{2}{{\left( {1 - w} \right)\left( {1 + w} \right)}}dw} + \int {\dfrac{w}{{1 - {w^2}}}dw} = \int {\dfrac{1}{u}du}$
Now using a partial fraction decomposition, we can write
$\dfrac{2}{{\left( {1 - w} \right)\left( {1 + w} \right)}} = \dfrac{1}{{w + 1}} - \dfrac{1}{{w - 1}}$
Therefore, we get
$\Rightarrow \int {\left( {\dfrac{1}{{\left( {w + 1} \right)}} + \dfrac{1}{{\left( {w - 1} \right)}}} \right)dw} + \int {\dfrac{w}{{1 - {w^2}}}dw} = \int {\dfrac{1}{u}du}$
Now we know that
$\int {\dfrac{1}{x}dx} = \ln |x|$ and $\int {\dfrac{x}{{1 - {x^2}}}dx} = - \dfrac{1}{2}\ln |{x^2} - 1|$
Therefore, we get
$\ln |w + 1| - \ln |w - 1| - \dfrac{1}{2}\ln |{w^2} - 1| = \ln |u| + \ln |c|$
We know that
$\ln |a| - \ln |b| = \ln \dfrac{{|a|}}{{|b|}}$
$\ln |a| + \ln |b| = \ln |ab|$
$a\ln |b| = \ln {\left( b \right)^a}$
So, we get
$\ln \dfrac{{|w + 1|}}{{|w - 1|}} - \ln \sqrt {{w^2} - 1} = \ln |Cu|$
$\Rightarrow \ln \left( {\dfrac{{|w + 1|}}{{|w - 1|\sqrt {{w^2} - 1} }}} \right) = \ln |Cu|$
$\Rightarrow \dfrac{{|w + 1|}}{{|w - 1|\sqrt {{w^2} - 1} }} = |Cu|$
On squaring both sides, we get
$\Rightarrow \dfrac{{{{\left( {w + 1} \right)}^2}}}{{{{\left( {w - 1} \right)}^2}\left( {{w^2} - 1} \right)}} = {C^2}{u^2}$
$\Rightarrow \dfrac{{{{\left( {w + 1} \right)}^2}}}{{{{\left( {w - 1} \right)}^2}\left( {w + 1} \right)\left( {w - 1} \right)}} = {C^2}{u^2}$
On cancelling $\left( {w + 1} \right)$ from both numerator and denominator, we get
$\Rightarrow \dfrac{{\left( {w + 1} \right)}}{{{{\left( {w - 1} \right)}^3}}} = {C^2}{u^2}$
$\Rightarrow \dfrac{{\left( {w + 1} \right)}}{{{{\left( {w - 1} \right)}^3}}} = {C^2}{u^2}$
$\Rightarrow \left( {w + 1} \right) = a{u^2}{\left( {w - 1} \right)^3}$ where $a = {C^2}$
Now, restoring the earlier $w$ substitution, using $w = \dfrac{v}{u}$ we get,
$\dfrac{v}{u} + 1 = a{u^2}{\left( {\dfrac{v}{u} - 1} \right)^3}$
Taking L.C.M we get
$\dfrac{{v + u}}{u} = a{u^2}{\left( {\dfrac{{v - u}}{u}} \right)^3}$
$\Rightarrow \dfrac{{v + u}}{u} = a{u^2}\dfrac{{{{\left( {v - u} \right)}^3}}}{{{u^3}}}$
$\Rightarrow v + u = a{\left( {v - u} \right)^3}$
Now, we will finally restore the earlier substitutions for $u$ and $v$ using equation $\left( 4 \right)$ and $\left( 5 \right)$ we get
$\left( {y - 1} \right) + \left( {x - 1} \right) = a{\left( {\left( {y - 1} \right) - \left( {x - 1} \right)} \right)^3}$
$\Rightarrow y + x - 2 = a{\left( {y - x} \right)^3}$
which is the required general solution of the given differential equation.

Note:
First order differential equations that can be written in the form $y' = F\left( {\dfrac{y}{x}} \right)$ are called Homogeneous differential equations. The given differential equation is also an example of Homogeneous differential equation. So, whenever you get these types of equations, always try to use the concept of substitution method. The basic idea behind this method is to transform and manipulate differential equations and solve them. The key idea is to replace the dependent variable or independent variable by a new variable that is expressed in terms of both of them.