4 Change of variables and chain rule in differential equation

These are examples doing change of variable for an ode.

4.1 Example 1 Change of the independent variable using \(z=g\left ( x\right ) \)

Given the ode

\[ \frac {d^{2}y}{dx^{2}}+\frac {dy}{dx}+y=\sin \left ( x\right ) \]

And we are asked to do change of variables from \(x\) to \(z\) where \(z=g\left ( x\right ) \). In this, we can also write

\[ x=g^{-1}\left ( z\right ) \]

Where \(g^{-1}\left ( z\right ) \) is the inverse function. Using chain rule gives

\[ \frac {dy}{dx}=\frac {dy}{dz}\frac {dz}{dx}\]

And for second derivative

\begin{align*} \frac {d^{2}y}{dx^{2}} & =\frac {d}{dx}\left ( \frac {dy}{dx}\right ) \\ & =\frac {d}{dx}\left ( \frac {dy}{dz}\frac {dz}{dx}\right ) \end{align*}

And now we use the product rule, which is \(\frac {d}{dx}\left ( ab\right ) =a^{\prime }b+ab^{\prime }\) on the above, which gives

\begin{equation} \frac {d^{2}y}{dx^{2}}=\left ( \frac {d}{dx}\frac {dy}{dz}\right ) \left ( \frac {dz}{dx}\right ) +\left ( \frac {dy}{dz}\right ) \left ( \frac {d}{dx}\frac {dz}{dx}\right ) \tag {1}\end{equation}

Let us do each of the terms on the right above one by one.  The second term on the RHS above is easy. It is

\begin{equation} \left ( \frac {dy}{dz}\right ) \left ( \frac {d}{dx}\frac {dz}{dx}\right ) =\left ( \frac {dy}{dz}\right ) \left ( \frac {d^{2}z}{dx^{2}}\right ) \tag {2}\end{equation}

It is the first term in (1) which needs more care. The problem is how to handle \(\frac {d}{dx}\frac {dy}{dz}\)? Since the denominators are different. The trick is to write \(\frac {d}{dx}\frac {dy}{dz}\) as \(\frac {d}{dz}\frac {dz}{dx}\left ( \frac {dy}{dz}\right ) \) which does not change anything, but now we can change the order and write this as \(\frac {dz}{dx}\frac {d}{dz}\left ( \frac {dy}{dz}\right ) \) which now makes the denominator the same and now it is free sailing:

\begin{align*} \frac {d}{dx}\frac {dy}{dz} & =\frac {d}{dz}\frac {dz}{dx}\left ( \frac {dy}{dz}\right ) \\ & =\frac {dz}{dx}\frac {d}{dz}\left ( \frac {dy}{dz}\right ) \\ & =\frac {dz}{dx}\left ( \frac {d^{2}y}{dz^{2}}\right ) \end{align*}

Therefore, the first term in (1) becomes

\begin{align} \left ( \frac {d}{dx}\frac {dy}{dz}\right ) \left ( \frac {dz}{dx}\right ) & =\frac {dz}{dx}\left ( \frac {d^{2}y}{dz^{2}}\right ) \left ( \frac {dz}{dx}\right ) \nonumber \\ & =\left ( \frac {dz}{dx}\right ) ^{2}\left ( \frac {d^{2}y}{dz^{2}}\right ) \tag {3}\end{align}

Using (2,3) then we have

\[ \frac {d^{2}y}{dx^{2}}=\left ( \frac {dz}{dx}\right ) ^{2}\left ( \frac {d^{2}y}{dz^{2}}\right ) +\left ( \frac {dy}{dz}\right ) \left ( \frac {d^{2}z}{dx^{2}}\right ) \]

Hence the original ode now becomes

\begin{align*} \frac {d^{2}y}{dx^{2}}+\frac {dy}{dx}+y & =\sin \left ( x\right ) \\ \overset {y^{\prime \prime }\left ( x\right ) }{\overbrace {\left ( \frac {dz}{dx}\right ) ^{2}\left ( \frac {d^{2}y}{dz^{2}}\right ) +\left ( \frac {dy}{dz}\right ) \left ( \frac {d^{2}z}{dx^{2}}\right ) }}+\overset {y^{\prime }\left ( x\right ) }{\overbrace {\frac {dy}{dz}\frac {dz}{dx}}}+y\left ( z\right ) & =\sin \left ( g^{-1}\left ( z\right ) \right ) \end{align*}

We could have written the RHS above as just \(\sin \left ( x\right ) \) instead of \(\sin \left ( g^{-1}\left ( z\right ) \right ) \) but since the independent variable is now \(z\), this seemed better to do it this way. But both are correct.  Now, since \(z=g\left ( x\right ) \) the above can also be written as

\begin{align*} \left ( \frac {dg}{dx}\right ) ^{2}\left ( \frac {d^{2}y}{dz^{2}}\right ) +\left ( \frac {dy}{dz}\right ) \left ( \frac {d^{2}g}{dx^{2}}\right ) +\frac {dy}{dz}\frac {dg}{dx}+y\left ( z\right ) & =\sin \left ( g^{-1}\left ( z\right ) \right ) \\ \left ( g^{\prime }\left ( x\right ) \right ) ^{2}y^{\prime \prime }\left ( x\right ) +y^{\prime }\left ( z\right ) g^{\prime \prime }\left ( x\right ) +y^{\prime }\left ( z\right ) g^{\prime }\left ( x\right ) +y\left ( z\right ) & =\sin \left ( x\right ) \end{align*}

OK, since the above was so much fun, lets do third derivative \(\frac {d^{3}y}{dx^{3}}\)

\begin{align} \frac {d^{3}y}{dx^{3}} & =\frac {d}{dx}\left ( \frac {d^{2}y}{dx^{2}}\right ) \nonumber \\ & =\frac {d}{dx}\left ( \left ( \frac {dz}{dx}\right ) ^{2}\left ( \frac {d^{2}y}{dz^{2}}\right ) +\left ( \frac {dy}{dz}\right ) \left ( \frac {d^{2}z}{dx^{2}}\right ) \right ) \nonumber \\ & =\frac {d}{dx}\left [ \left ( \frac {dz}{dx}\right ) ^{2}\left ( \frac {d^{2}y}{dz^{2}}\right ) \right ] +\frac {d}{dx}\left [ \left ( \frac {dy}{dz}\right ) \left ( \frac {d^{2}z}{dx^{2}}\right ) \right ] \tag {4}\end{align}

Each term above is now found. Looking at first term in (4)

\[ \frac {d}{dx}\left [ \left ( \frac {dz}{dx}\right ) ^{2}\left ( \frac {d^{2}y}{dz^{2}}\right ) \right ] \]

Using the product rule, which is \(\frac {d}{dx}\left ( ab\right ) =a^{\prime }b+ab^{\prime }\) on the above gives

\[ \frac {d}{dx}\left [ \left ( \frac {dz}{dx}\right ) ^{2}\left ( \frac {d^{2}y}{dz^{2}}\right ) \right ] =\frac {d}{dx}\left [ \left ( \frac {dz}{dx}\right ) ^{2}\right ] \frac {d^{2}y}{dz^{2}}+\left ( \frac {dz}{dx}\right ) ^{2}\frac {d}{dx}\left [ \frac {d^{2}y}{dz^{2}}\right ] \]

But \(\frac {d}{dx}\left [ \left ( \frac {dz}{dx}\right ) ^{2}\right ] =2\frac {dz}{dx}\frac {d^{2}z}{dx}\) and for \(\frac {d}{dx}\left ( \frac {d^{2}y}{dz^{2}}\right ) \) we have to use the same trick as before by writing \(\frac {d}{dx}\left ( \frac {d^{2}y}{dz^{2}}\right ) =\frac {d}{dz}\frac {dz}{dx}\left ( \frac {d^{2}y}{dz^{2}}\right ) =\frac {dz}{dx}\frac {d}{dz}\left ( \frac {d^{2}y}{dz^{2}}\right ) \) and now we have \(\frac {d}{dx}\left ( \frac {d^{2}y}{dz^{2}}\right ) =\frac {dz}{dx}\frac {d^{3}y}{dz^{3}}\). Hence the first term in (4) is now done.

\begin{align} \frac {d}{dx}\left [ \left ( \frac {dz}{dx}\right ) ^{2}\left ( \frac {d^{2}y}{dz^{2}}\right ) \right ] & =2\frac {dz}{dx}\frac {d^{2}z}{dx^{2}}\frac {d^{2}y}{dz^{2}}+\left ( \frac {dz}{dx}\right ) ^{2}\frac {dz}{dx}\frac {d^{3}y}{dz^{3}}\nonumber \\ & =2\frac {dz}{dx}\frac {d^{2}z}{dx^{2}}\frac {d^{2}y}{dz^{2}}+\left ( \frac {dz}{dx}\right ) ^{3}\frac {d^{3}y}{dz^{3}} \tag {5}\end{align}

Now we look at the second term in (4) which is \(\frac {d}{dx}\left [ \left ( \frac {dy}{dz}\right ) \left ( \frac {d^{2}z}{dx^{2}}\right ) \right ] \) and apply the product rule, this gives

\begin{align} \frac {d}{dx}\left [ \left ( \frac {dy}{dz}\right ) \left ( \frac {d^{2}z}{dx^{2}}\right ) \right ] & =\frac {d}{dx}\left [ \frac {dy}{dz}\right ] \left ( \frac {d^{2}z}{dx^{2}}\right ) +\frac {dy}{dz}\frac {d}{dx}\left [ \frac {d^{2}z}{dx^{2}}\right ] \nonumber \\ & =\frac {d}{dz}\frac {dz}{dx}\left [ \frac {dy}{dz}\right ] \left ( \frac {d^{2}z}{dx^{2}}\right ) +\frac {dy}{dz}\frac {d^{3}z}{dx^{3}}\nonumber \\ & =\frac {dz}{dx}\frac {d}{dz}\left [ \frac {dy}{dz}\right ] \left ( \frac {d^{2}z}{dx^{2}}\right ) +\frac {dy}{dz}\frac {d^{3}z}{dx^{3}}\nonumber \\ & =\frac {dz}{dx}\frac {d^{2}y}{dz^{2}}\left ( \frac {d^{2}z}{dx^{2}}\right ) +\frac {dy}{dz}\frac {d^{3}z}{dx^{3}} \tag {6}\end{align}

That is it. We are done. (5,6) are the two terms in (4). Therefore

\begin{align*} \frac {d^{3}y}{dx^{3}} & =2\frac {dz}{dx}\frac {d^{2}z}{dx^{2}}\frac {d^{2}y}{dz^{2}}+\left ( \frac {dz}{dx}\right ) ^{3}\frac {d^{3}y}{dz^{3}}+\frac {dz}{dx}\frac {d^{2}y}{dz^{2}}\left ( \frac {d^{2}z}{dx^{2}}\right ) +\frac {dy}{dz}\frac {d^{3}z}{dx^{3}}\\ & =3\frac {dz}{dx}\frac {d^{2}z}{dx^{2}}\frac {d^{2}y}{dz^{2}}+\left ( \frac {dz}{dx}\right ) ^{3}\frac {d^{3}y}{dz^{3}}+\frac {dy}{dz}\frac {d^{3}z}{dx^{3}}\end{align*}

Now, since \(z=g\left ( x\right ) \) the above can also be written as

\[ y^{\prime \prime \prime }\left ( x\right ) =3g^{\prime }\left ( x\right ) g^{\prime \prime }\left ( x\right ) y^{\prime \prime }\left ( z\right ) +\left ( g^{\prime }\left ( x\right ) \right ) ^{3}y^{\prime \prime \prime }\left ( z\right ) +y^{\prime }\left ( z\right ) g^{\prime \prime \prime }\left ( x\right ) \]

This table show summary of transformation for each derivative \(y^{\left ( n\right ) }\left ( x\right ) \) when using change of variables \(z=g\left ( x\right ) \)

\(y^{\prime }\left ( x\right ) \)	\(y^{\prime }\left ( z\right ) g^{\prime }\left ( x\right ) \)
\(y^{\prime \prime }\left ( x\right ) \)	\(\left ( g^{\prime }\left ( x\right ) \right ) ^{2}y^{\prime \prime }\left ( z\right ) +y^{\prime }\left ( z\right ) g^{\prime \prime }\left ( x\right ) \)
\(y^{\prime \prime \prime }\left ( x\right ) \)	\(3g^{\prime }\left ( x\right ) g^{\prime \prime }\left ( x\right ) y^{\prime \prime }\left ( z\right ) +\left ( g^{\prime }\left ( x\right ) \right ) ^{3}y^{\prime \prime \prime }\left ( z\right ) +y^{\prime }\left ( z\right ) g^{\prime \prime \prime }\left ( x\right ) \)
\(y^{\prime \prime \prime \prime }\left ( x\right ) \)	\(3\left ( g^{\prime \prime }\left ( x\right ) \right ) ^{2}y^{\prime \prime }\left ( z\right ) +4g^{\prime }\left ( x\right ) y^{\prime \prime }\left ( z\right ) g^{\prime \prime \prime }\left ( x\right ) +6\left ( g^{\prime }\left ( x\right ) \right ) ^{2}g^{\prime \prime }\left ( x\right ) y^{\prime \prime \prime }\left ( z\right ) +y^{\prime }\left ( z\right ) g^{\prime \prime \prime \prime }\left ( x\right ) +\left ( g^{\prime }\left ( x\right ) \right ) ^{4}y^{\prime \prime \prime \prime }\left ( z\right ) \)

Strictly speaking, it would be better to use different variable than \(y\) when changing the independent variable. i.e. instead of writing \(y\left ( z\right ) \) in all the above, we should write \(u\left ( z\right ) \) in its place. So the above table will look like

\(y^{\prime }\left ( x\right ) \)	\(u^{\prime }\left ( z\right ) g^{\prime }\left ( x\right ) \)
\(y^{\prime \prime }\left ( x\right ) \)	\(\left ( g^{\prime }\left ( x\right ) \right ) ^{2}u^{\prime \prime }\left ( z\right ) +u^{\prime }\left ( z\right ) g^{\prime \prime }\left ( x\right ) \)
\(y^{\prime \prime \prime }\left ( x\right ) \)	\(3g^{\prime }\left ( x\right ) g^{\prime \prime }\left ( x\right ) u^{\prime \prime }\left ( z\right ) +\left ( g^{\prime }\left ( x\right ) \right ) ^{3}u^{\prime \prime \prime }\left ( z\right ) +u^{\prime }\left ( z\right ) g^{\prime \prime \prime }\left ( x\right ) \)
\(y^{\prime \prime \prime \prime }\left ( x\right ) \)	\(3\left ( g^{\prime \prime }\left ( x\right ) \right ) ^{2}u^{\prime \prime }\left ( z\right ) +4g^{\prime }\left ( x\right ) u^{\prime \prime }\left ( z\right ) g^{\prime \prime \prime }\left ( x\right ) +6\left ( g^{\prime }\left ( x\right ) \right ) ^{2}g^{\prime \prime }\left ( x\right ) u^{\prime \prime \prime }\left ( z\right ) +y^{\prime }\left ( z\right ) g^{\prime \prime \prime \prime }\left ( x\right ) +\left ( g^{\prime }\left ( x\right ) \right ) ^{4}u^{\prime \prime \prime \prime }\left ( z\right ) \)

So any place where \(y\left ( z\right ) \) shows in the transformed expression, it should be written with new letter for the dependent variable \(u\left ( z\right ) \). But this is not always enforced.

4.2 Example 2 Change of the independent variable using \(t=\ln \left ( x\right ) \) Euler ode

Given the ode

\[ x^{2}\frac {d^{2}y}{dx^{2}}+2x\frac {dy}{dx}+y=0 \]

And asked to do change of variable \(t=\ln \left ( x\right ) \)

\begin{align*} \frac {dy}{dx} & =\frac {dy}{dt}\frac {dt}{dx}\\ & =\frac {dy}{dt}\frac {1}{x}\end{align*}

And

\begin{align*} \frac {d^{2}y}{dx^{2}} & =\frac {d}{dx}\left ( \frac {dy}{dx}\right ) \\ & =\frac {d}{dx}\left ( \frac {dy}{dt}\frac {1}{x}\right ) \\ & =\frac {d}{dx}\left [ \frac {dy}{dt}\right ] \frac {1}{x}+\frac {dy}{dt}\frac {d}{dx}\left ( \frac {1}{x}\right ) \\ & =\frac {d}{dt}\frac {dt}{dx}\left [ \frac {dy}{dt}\right ] \frac {1}{x}-\frac {dy}{dt}\frac {1}{x^{2}}\\ & =\frac {dt}{dx}\frac {d^{2}y}{dt^{2}}\frac {1}{x}-\frac {dy}{dt}\frac {1}{x^{2}}\\ & =\frac {1}{x}\frac {d^{2}y}{dt^{2}}\frac {1}{x}-\frac {dy}{dt}\frac {1}{x^{2}}\\ & =\frac {1}{x^{2}}\frac {d^{2}y}{dt^{2}}-\frac {dy}{dt}\frac {1}{x^{2}}\end{align*}

Hence the original ode becomes

\begin{align*} x^{2}\left ( \frac {1}{x^{2}}\frac {d^{2}y}{dt^{2}}-\frac {dy}{dt}\frac {1}{x^{2}}\right ) +2x\left ( \frac {dy}{dt}\frac {1}{x}\right ) +y & =0\\ \frac {d^{2}y}{dt^{2}}-\frac {dy}{dt}+2\frac {dy}{dt}+y & =0\\ \frac {d^{2}y}{dt^{2}}+\frac {dy}{dt}+y & =0 \end{align*}

4.3 Example 3 Change of the dependent variable using \(y=x^{r}\) Euler ode

Given the ode

\[ x^{2}\frac {d^{2}y}{dx^{2}}+2x\frac {dy}{dx}+y=0 \]

And asked to do change of variable \(y=x^{r}\)

\[ \frac {dy}{dx}=rx^{r-1}\]

And

\begin{align*} \frac {d^{2}y}{dx^{2}} & =\frac {d}{dx}\left ( rx^{r-1}\right ) \\ & =r\left ( r-1\right ) x^{r-2}\end{align*}

Hence the original ode becomes

\begin{align*} x^{2}\left ( r\left ( r-1\right ) x^{r-2}\right ) +2x\left ( rx^{r-1}\right ) +x^{r} & =0\\ r\left ( r-1\right ) x^{r}+2rx^{r}+x^{r} & =0\\ r\left ( r-1\right ) +2r+1 & =0 \end{align*}

Solving for \(r\) gives the roots. Hence solutions are \(y_{1}=x^{r_{1}}\) and \(y_{2}=x^{r_{2}}\). Final solution is therefore

\begin{align*} y & =c_{1}y_{1}+c_{2}y_{2}\\ & =c_{1}x^{r_{1}}+c_{2}x^{r_{2}}\end{align*}

This method of solving the Euler ode is much simpler than using \(t=\ln \left ( x\right ) \) change of variables but for some reason most text books use the later one.