Appropriately Applying the Chain Rule to a Substitution made to an ODE

Question

I have the differential equation, $$\frac{dG(-5x)}{dx}=-\frac{1}{6}\left(g_1(x)+\frac{dg_0(x)}{dx}\right).$$ If I make the substitution $u=-5x$, does the above differential equation collapse to form $$\frac{dG}{du}=-\frac{1}{6}\left(g_1\left(\frac{-u}{5}\right)+-5\frac{dg_0}{du}\right),$$ as $$\frac{dg_0}{dx}=\frac{dg_0}{du}\frac{du}{dx}.$$ I am unsure of why the term on the LHS doesn't change. Thank you a lot in advance.

Answer 1

It could be that your equation is correctly written (at present) as $$\frac{dG(-5x)}{dz}=-\frac{1}{6}\left(g_1(x)+\frac{dg_0(x)}{dx}\right),$$ but I suspect a typo. I think your equation is supposed to be $$\frac{dG(-5x)}{dx}=-\frac{1}{6}\left(g_1(x)+\frac{dg_0(x)}{dx}\right),$$ instead. The left-hand side of this equation could be interpreted in two ways that I can think of. One, as $$\frac{d}{dx}\bigl[G(-5x)\bigr],\tag{1}$$ the other, as simply $$G'(-5x).\tag{2}$$ If $(1)$ is intended, as I suspect, then the Chain Rule lets us rewrite the left-hand side as $$\frac{d}{dx}\bigl[G(-5x)\bigr]=\frac{d}{dy}\bigl[G(y)\bigr]_{y=-5x}\cdot\frac{d}{dx}\bigl[-5x\bigr]=-5G'(-5x),$$ so the equation becomes $$-5G'(-5x)=-\frac{1}{6}\left(g_1(x)+g_0'(x)\right),$$ or (after substitution) $$-5G'(u)=-\frac{1}{6}\left(g_1\left(-\frac{u}{5}\right)+g_0'\left(-\frac{u}{5}\right)\right).$$ But why wouldn't the Chain rule apply to the differentiation on the right-hand side? Well, it does, but here's how it does under interpretation $(1).$ $$\begin{eqnarray}\frac{dg_0(x)}{dx}&=&\frac{dg_0\left(-\frac{u}{5}\right)}{dx}\\&=&\frac{d}{dx}\left[g_0\left(-\frac{u}{5}\right)\right]\\&=&\frac{d}{dy}\bigl[g_0(y)\bigr]_{y=-\frac{u}{5}}\cdot\frac{d}{du}\left[-\frac{u}{5}\right]\cdot\frac{du}{dx}\\&=&g_0'\left(-\frac{u}5\right)\cdot-\frac15\cdot-5\\&=&g_0'\left(-\frac{u}5\right).\end{eqnarray}$$

If $(2)$ is intended, then the substitution is even more straightforward, and yields $$G'(u)=-\frac{1}{6}\left(g_1\left(-\frac{u}{5}\right)+g_0'\left(-\frac{u}{5}\right)\right),$$ and the Chain Rule goes completely un-invoked.