Question

Solve

$\frac{dy}{dx}$ =(x+2y-3)/(2x+y-3)

Answer 1

Our standard toolkit for DE's cannot be used. However, we can perform a transformation to remove the constants from the linear numerator and denominator.

Consider the simultaneous equations

$\begin{cases} x+2y-3=0 \\ 2x+y-3=0 \end{cases} \implies \begin{cases} x=1\\ y=1 \end{cases}$

As a result we perform two linear transformations:

$\begin{cases} u=x-1 \\ v=y-1 \end{cases} \implies \begin{cases} x=u+1\\ y=v+1 \end{cases} \implies \begin{cases} \frac{dx}{du}=1\\ \frac{dy}{dv}=1 \end{cases}$

And if we substitute into the DE we get

$\frac{dy}{dx}=\frac{(u+1)+2(v+1)-3}{2(u+1)+2(v+1)-3}\\ \frac{dy}{dx}=\frac{u+1+2v+2-3}{2u+2+v+1-3}\\ \frac{dy}{dx}=\frac{u+2v}{2u+v}$

And utilising the chain rule we have

$\frac{dy}{dx}= \frac{dy}{dv}\frac{dv}{du} \frac{du}{dx} \implies \frac{dy}{dx}= \frac{dv}{du}$

Thus we have a transformed equation

$\frac{dv}{du}= \frac{u+2v}{2u+v}\\$

This is now in a form that we can handle using a substitution of form v =wu, which if we differentiate wrt u using the product gives us.

Using this substitution into our modified DE, we get

$u \frac{dw}{du}= \frac{1+w^2}{2+w}\\$

This is now a separable DE, so we can rearrange and separate the variables to get:

$\int \frac{1}{u}du = \int \frac{2+w}{1+w^2}\\ \ln | w + 1 | − \ln | w − 1 | − 1 2 \ln ∣ ∣ w 2 − 1 ∣ ∣ = \ln | u | + \ln C$

$\frac{| w + 1 |}{ | w − 1 | √ w 2 − 1} = | C u |\\ \therefore (w+1)=Au^2(w-1)^3$

Then restoring the earlier w substitution, using $w = \frac{v}{ u}$  we get:

$∴ v + u = A ( v − u )^ 3$

$( y − 1 ) + ( x − 1 ) = A ( ( y − 1 ) − ( x − 1 ) )^ 3$

$∴ y + x − 2 = A ( y − x )^ 3$

This is the General Solution, in implicit form.