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Consider the equation $$\frac{dy}{dx}=F(\frac{ax+by+c}{dx+ey+f})$$

Show that if $ae \neq bd$ then there exists constants $h \; , \; k$ such that the substitution $x=z-h$ and $y=w-k$ converts the above equation into a homogenous equation.

I'm quite new to differential equations . Can you please provide some idea to proceed?

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Case1 : $ae - bd\ne0$, Put $ x=z-h,y=w-k\Rightarrow\frac{dw}{dz}=\frac{dy}{dx}$ using all these in the given equation we have,

$\frac{dw}{dz}=F\left(\frac{a(z-h)+b(w-k)+c}{d(z-h)+e(w-k)+f}\right)=F\left(\frac{az+bw-(ah+bk-c)}{dz+ew-(dh+ek-f)}\right) =F\left(\frac {az+bw}{dz+ew}\right)$

This relation will hold if we take, $$ah+bk-c=0,dh+ek-f=0$$ which are linear equation in $h,k$ and can be solved for $h,k$

$\frac{dw}{dz}=F\left(\frac {az+bw}{dz+ew}\right)=F\left(\frac {a+b\left(\frac{w}{z}\right)}{d+e\left(\frac{w}{z}\right)}\right)=G\left(\frac{w}{z}\right)$, which is the required form.

Case2 : $ae - bd=0$, If you need to know this case, ask for explanation

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Note that $ae - bd = \det \begin{pmatrix} a & b \\ d & e\end{pmatrix}$. If $ae \neq bd$, then we can solve $h,k$ (uniquely) for $$\begin{pmatrix} a & b \\ d & e\end{pmatrix} \begin{pmatrix} h \\ k\end{pmatrix} = \begin{pmatrix}c \\ f\end{pmatrix}$$ and therefore $$ax+by+c = a(z-h)+b(w-k)+c=az+bw-(ah+bk-c) = az+bw$$ $$dx+ey+f = d(z-h)+e(w-k)+f=dz+ew-(dh+ek-f) = dz+ew$$

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Hint:
Substitute $x=z-h,y=w-k$ into $\dfrac{dy}{dx}=F\left(\dfrac{ax+by+c}{dx+ey+f}\right)$. What is the relation between $\dfrac{dy}{dx}$ and $\dfrac{dw}{dz}$? Is the RHS homogeneous (with respect to $w$ and $z$)? How to choose $h$ and $k$?

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