Q.The solution of $(x+y)^2\dfrac{dy}{dx}=1$ is... (A) $x=\tan^{-1}(x+y)+c$ (B) $y\tan^{-1}\dfrac{x}{y}=c$ (C) $y=\tan^{-1}(x+y)+c$ (D) $y+\tan^{-1}(x+y)=c$

A function $f(x,y)$ is a homogeneous function of degree $n$ if $f(tx,ty)=t^n f(x,y)$ for every suitably restricted $x,y,t$ (Euler's homogeneity). A homogeneous function of degree zero can always be written purely as a function of the single ratio $\dfrac{y}{x}$ (or $\dfrac{x}{y}$): $f(x,y)=g\!\left(\dfrac{y}{x}\right)$.

Homogeneous differential equation. An ODE is in homogeneous form if it can be written as

Equivalently, $M(x,y)\,dx+N(x,y)\,dy=0$ is homogeneous exactly when $M$ and $N$ are homogeneous functions of the same degree — because then $f(x,y)=-M/N$ is automatically homogeneous of degree $0$. (This use of the word "homogeneous" for the equation is a different meaning from calling the constant term $g(x)=0$ in a linear equation "homogeneous" — Definition 10.7 versus Definition 10.12 in the textbook — so the two uses should not be confused.)

Solution method (Theorem 10.1). Substitute $y=vx$ (so $v=\dfrac{y}{x}$), giving $\dfrac{dy}{dx}=v+x\dfrac{dv}{dx}$. The homogeneous equation becomes

which is variables-separable in $v$ and $x$:

Integrate both sides, then replace $v$ by $\dfrac{y}{x}$ to return to the original variables. …