Substitution Method

When an equation has the shape

— a function of the single linear combination $ax+by+c$, rather than of $x$ and $y$ separately — it is not immediately separable, but a linear substitution converts it into an equation that is.

Method.

1. If $a\ne0$ and $b\ne0$: substitute $z=ax+by+c$. Then $\dfrac{dz}{dx}=a+b\dfrac{dy}{dx}$, so $\dfrac{dy}{dx}=\dfrac{1}{b}\left(\dfrac{dz}{dx}-a\right)$. Substituting into the original equation gives an equation relating $z$ and $x$ alone, which reduces the given equation to variables-separable form; solve for $z(x)$, then replace $z$ by $ax+by+c$ to recover the answer in $x$ and $y$.
2. If $a=0$ or $b=0$, the equation is already separable in $x$ and $y$ directly, and no substitution is needed. Worked pattern (Example 10.13). $y'=\sin^2(x-y+1)$: put $z=x-y+1$, so $\dfrac{dz}{dx}=1-\dfrac{dy}{dx}$, giving $1-\dfrac{dz}{dx}=\sin^2z$, i.e. $\dfrac{dz}{dx}=1-\sin^2z=\cos^2z$. Separating: $\sec^2z\,dz=dx$, integrating to $\tan z=x+C$, i.e. $\tan(x-y+1)=x+C$. Worked pattern (Example 10.14, a fractional right side). $\dfrac{dy}{dx}=\dfrac{4x+y-1}{2x+y}$: putting $z=4x+2y-1$ (chosen to match the denominator's structure after scaling) reduces the equation to $\dfrac{dz}{dx}=4+\dfrac{2z}{z}$-type separable form; integrating (using the substitution $z=u^2$ to handle the resulting square-root term) and replacing $z$ back gives the implicit general solution in $x,y$. …