Differential equations – Ordinary differential equations – Euler differential equations

Definition: A linear differential equation of the form

\[t^2y''(t)+\alpha ty'(t)+\beta y(t)=0,\quad\alpha,\beta\in\mathbb{R},\quad t>0\]

is called an Euler differential equation.

Using the substitution \(t=e^x\) or equivalently \(x=\ln(t)\) we obtain:

\[\frac{dy}{dt}=\frac{dy}{dx}\cdot\frac{dx}{dt}=\frac{1}{t}\cdot\frac{dy}{dx}\quad\Longrightarrow\quad t\,\frac{dy}{dt}=\frac{dy}{dx}\]

and

\[\frac{d^2y}{dt^2}=\frac{d}{dt}\left(\frac{1}{t}\cdot\frac{dy}{dx}\right)=\frac{1}{t}\cdot\frac{d^2y}{dx^2}\cdot\frac{dx}{dt}-\frac{1}{t^2}\cdot\frac{dy}{dx} =\frac{1}{t^2}\left(\frac{d^2y}{dx^2}-\frac{dy}{dx}\right)\quad\Longrightarrow\quad t^2\,\frac{d^2y}{dt^2}=\frac{d^2y}{dx^2}-\frac{dy}{dx}.\]

With this the differential equation \(t^2y''(t)+\alpha ty'(t)+\beta y(t)=0\) transforms into \(Y''(x)+(\alpha-1)Y'(x)+\beta Y(x)=0\), where \(Y(x)=y(t)\).
This is a second-order linear differential equation with constant coefficients and characteristic equation \(r^2+(\alpha-1)r+\beta=0\). This is obtained by trying to find a solution of the form \(Y(x)=e^{rx}\). That is similar to trying to find a solution of the form \(y(t)=t^r\) of the original Euler differential equation:

\[r(r-1)+\alpha r+\beta=0\quad\Longleftrightarrow\quad r^2+(\alpha-1)r+\beta=0.\]

We consider three possibilities for the discriminant \(D=(\alpha-1)^2-4\beta\): \(D>0\), \(D=0\) and \(D<0\).

If \(D>0\) then the characteristic equation has two different real solutions, say \(r_1\) and \(r_2\).
Then the general solution is \(y(t)=c_1t^{r_1}+c_2t^{r_2}\).

If \(D=0\) then the characteristic equation has two equal real solutions \(r_1=r_2=r\). Then we have using \(x=\ln(t)\) that \(xe^{r x}=t^r\ln(t)\).
Then the general solution is: \(y(t)=c_1t^r+c_2t^r\ln(t)\).

If \(D<0\) then the characteristic equation has two nonreal solutions, say \(r_{1,2}=\gamma\pm i\delta\) with \(\delta\neq0\). Then we have using \(x=\ln(t)\) that \(e^{\gamma x}\cos(\beta x)=t^{\gamma}\cos(\beta\ln(t))\) and \(e^{\gamma x}\sin(\beta x)=t^{\gamma}\sin(\beta\ln(t))\).
Then the general solution is \(y(t)=c_1t^{\gamma}\cos(\beta\ln(t))+c_2t^{\gamma}\sin(\beta\ln(t))\).

Hence: if \(r_1\) and \(r_2\) are the two solutions of the characteristic equation, then

1. if \(r_1,r_2\in\mathbb{R}\) with \(r_1\neq r_2\), then: \(y(t)=c_1t^{r_1}+c_2t^{r_2}\) with \(c_1,c_2\in\mathbb{R}\),


2. if \(r_1,r_2\in\mathbb{R}\) with \(r_1=r_2=r\), then: \(y(t)=c_1t^{r}+c_2t^{r}\ln(t)\) with \(c_1,c_2\in\mathbb{R}\),


3. if \(r_1,r_2\notin\mathbb{R}\), so \(r_{1,2}=\gamma\pm i\delta\) with \(\delta\neq0\), then: \(y(t)=c_1t^{\gamma}\cos(\delta\ln(t))+c_2t^{\gamma}\sin(\delta\ln(t))\) with \(c_1,c_2\in\mathbb{R}\).

Examples:

1) Consider \(t^2y''(t)-ty'(t)-3y(t)=0\) for \(t>0\), then the characteristic equation is: \(r(r-1)-r-3=0\) or equivalently \((r+1)(r-3)=0\).
Then the general solution is: \(y(t)=c_1t^3+c_2t^{-1}\).

2) Consider \(t^2y''(t)-3ty'(t)+4y(t)=0\) for \(t>0\), then the characteristic equation is: \(r(r-1)-3r+4=0\) or equivalently \((r-2)^2=0\).
Then the general solution is: \(y(t)=c_1t^2+c_2t^2\ln(t)\).

3) Consider \(t^2y''(t)-ty'(t)+5y(t)=0\) for \(t>0\), then the characteristic equation is: \(r(r-1)-r+5=0\) or equivalently \(((r-1)^2+4=0\).
Then the general solution is: \(y(t)=c_1t\cos(2\ln(t))+c_2t\sin(2\ln(t))\).

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Last modified on April 19, 2021