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Examples doing change of variable in differential equations

Nasser M. Abbasi

October 29, 2021   Compiled on October 29, 2021 at 3:32am

Contents

 1 Change of variable in the independent variable
  1.1 Example 1. Euler ODE
 2 Change of variable in the dependent variable
  2.1 Example 1. Bessel ODE

1 Change of variable in the independent variable

1.1 Example 1. Euler ODE

One way to solve Euler ODE \begin {equation} x^{2}y^{\prime \prime }\relax (x) +xy^{\prime }\relax (x) +y\left ( x\right ) =0\tag {1} \end {equation} is to do a change of variable in the independent variable.  Let \(x=e^{t}\). Then \begin {equation} \frac {dx}{dt}=e^{t}\tag {2} \end {equation} Also we have that \(\ln x=t\), which means \begin {equation} \frac {dt}{dx}=\frac {1}{x}\tag {3} \end {equation} To do this change of variable and obtain a new ode where now \(y\left ( x\right ) \) becomes \(y\relax (t) \), we start by changing \(y^{\prime }\relax (x) \) to \(y^{\prime }\relax (t) \) and changing \(y^{\prime \prime }\relax (x) \) to \(y^{\prime \prime }\relax (t) \). Given that\begin {equation} \frac {dy}{dx}=\frac {dy}{dt}\frac {dt}{dx}\tag {4} \end {equation} Substituting (3) into (4) gives\[ \frac {dy}{dx}=\frac {dy}{dt}\frac {1}{x}\] But \(\frac {1}{x}=e^{-t}\). The above becomes\begin {equation} \frac {dy}{dx}=e^{-t}\frac {dy}{dt}\tag {5} \end {equation} This was not too bad. Now we need to work on \(y^{\prime \prime }\left ( x\right ) \). \[ \frac {d^{2}y}{dx^{2}}=\frac {d}{dx}\left (\frac {dy}{dx}\right ) \] Substituting (5) into the above gives\[ \frac {d^{2}y}{dx^{2}}=\frac {d}{dx}\left (e^{-t}\frac {dy}{dt}\right ) \] Now here comes the important trick. We want \(\frac {d}{dx}\) to be \(\frac {d}{dt}\) in order to apply it on its argument. This is done by dividing the numerator and denominator of \(\frac {d}{dx}\) by \(dt\) which becomes\begin {align*} \frac {d^{2}y}{dx^{2}} & =\frac {\frac {d}{dt}}{\frac {dx}{dt}}\left ( e^{-t}\frac {dy}{dt}\right ) \\ & =\frac {dt}{dx}\frac {d}{dt}\left (e^{-t}\frac {dy}{dt}\right ) \end {align*}

But from (3) \(\frac {dt}{dx}=\frac {1}{x}=e^{-t}\). Hence the above becomes\[ \frac {d^{2}y}{dx^{2}}=e^{-t}\frac {d}{dt}\left (e^{-t}\frac {dy}{dt}\right ) \] Now we can apply the product rule to finish the above\begin {align} \frac {d^{2}y}{dx^{2}} & =e^{-t}\left (-e^{-t}\frac {dy}{dt}+e^{-t}\frac {d^{2}y}{dt^{2}}\right ) \nonumber \\ & =e^{-2t}\left (-\frac {dy}{dt}+\frac {d^{2}y}{dt^{2}}\right ) \nonumber \\ & =e^{-2t}\left (\frac {d^{2}y}{dt^{2}}-\frac {dy}{dt}\right ) \tag {6} \end {align}

We have now converted bot \(y^{\prime }\relax (x) \) and \(y^{\prime \prime }\relax (x) \,\). what is left is to plug these back into the original ODE (1). Which becomes\begin {align*} x^{2}y^{\prime \prime }\relax (x) +xy^{\prime }\relax (x) +y\left ( x\right ) & =0\\ x^{2}e^{-2t}\left (\frac {d^{2}y}{dt^{2}}-\frac {dy}{dt}\right ) +xe^{-t}\frac {dy}{dt}+y\relax (t) & =0 \end {align*}

But \(x=e^{t}\) and \(x^{2}=e^{2t}\). The above becomes\begin {align*} \frac {d^{2}y}{dt^{2}}-\frac {dy}{dt}+\frac {dy}{dt}+y\relax (t) & =0\\ \frac {d^{2}y}{dt^{2}}+y\relax (t) & =0 \end {align*}

This is now constant coefficient ODE. It is standard one which has the solution\[ y\relax (t) =A\cos \relax (t) +B\sin \relax (t) \] Now we switch back to \(x\). Since\(\ \ln x=t\), then the above becomes\[ y\relax (x) =A\cos \left (\ln x\right ) +B\sin \left (\ln x\right ) \] This completes the solution.

2 Change of variable in the dependent variable

2.1 Example 1. Bessel ODE

Given the ode\begin {equation} y^{\prime \prime }\relax (x) +\left (1-\frac {3}{4x^{2}}\right ) y\left ( x\right ) =0\tag {1} \end {equation} We will use the change of variable in the dependent variable \(y=ux^{\frac {1}{2}}\) to transform the above ODE to Bessel ODE\[ x^{2}u^{\prime \prime }+xu^{\prime }+\left (x^{2}-1\right ) u=0 \] Since \(y=ux^{\frac {1}{2}}\) then\begin {equation} \frac {dy}{dx}=\frac {du}{dx}x^{\frac {1}{2}}+u\frac {x^{\frac {=1}{2}}}{2}\tag {2} \end {equation} And\begin {align} \frac {d^{2}y}{dx^{2}} & =\frac {d}{dx}\left (\frac {du}{dx}x^{\frac {1}{2}}+u\frac {x^{\frac {=1}{2}}}{2}\right ) \nonumber \\ & =\frac {d}{dx}\left (\frac {du}{dx}x^{\frac {1}{2}}\right ) +\frac {d}{dx}\left (u\frac {x^{\frac {=1}{2}}}{2}\right ) \nonumber \\ & =\frac {d^{2}u}{dx^{2}}x^{\frac {1}{2}}+\frac {1}{2}\frac {du}{dx}x^{-\frac {1}{2}}+\frac {1}{2}\frac {du}{dx}x^{\frac {=1}{2}}-\frac {1}{4}ux^{-\frac {3}{2}}\nonumber \\ & =\frac {d^{2}u}{dx^{2}}x^{\frac {1}{2}}+\frac {du}{dx}x^{-\frac {1}{2}}-\frac {1}{4}ux^{-\frac {3}{2}}\tag {3} \end {align}

Substituting (2,3) into (1) gives\begin {align*} \frac {d^{2}u}{dx^{2}}x^{\frac {1}{2}}+\frac {du}{dx}x^{-\frac {1}{2}}-\frac {1}{4}ux^{-\frac {3}{2}}+\left (1-\frac {3}{4x^{2}}\right ) ux^{\frac {1}{2}} & =0\\ \frac {d^{2}u}{dx^{2}}x^{\frac {1}{2}}+\frac {du}{dx}x^{-\frac {1}{2}}-\frac {1}{4}ux^{-\frac {3}{2}}+ux^{\frac {1}{2}}-\frac {3}{4}ux^{-\frac {3}{2}} & =0\\ \frac {d^{2}u}{dx^{2}}x^{\frac {1}{2}}+\frac {du}{dx}x^{-\frac {1}{2}}-ux^{-\frac {3}{2}}+ux^{\frac {1}{2}} & =0 \end {align*}

Multiplying both side by \(x^{\frac {3}{2}}\) gives\begin {align*} x^{2}\frac {d^{2}u}{dx^{2}}+x\frac {du}{dx}-u+ux^{2} & =0\\ x^{2}\frac {d^{2}u}{dx^{2}}+x\frac {du}{dx}-\left (1-x^{2}\right ) u & =0\\ x^{2}\frac {d^{2}u}{dx^{2}}+x\frac {du}{dx}+\left (x^{2}-1\right ) u & =0 \end {align*}

Which is Bessel ode where \(v=1\).