# Holmes' question at Yahoo! Answers regarding a Cauchy-Euler equation

Staff member

#### MarkFL

Staff member
Hello Holmes,

Method 1:

A linear second order equation that can be expressed in the form:

$\displaystyle ax^2\frac{d^2y}{dx^2}+bx\frac{dy}{dx}+cy=h(x)$,

where $a$, $b$, and $c$ are constants is called a Cauchy-Euler equation.

Although work on this equation was published by Leonhard Euler in 1769 and later by Augustin Cauchy, its solution was known to John Bernoulli prior to 1700. These equations are also called equidimensional equations.

The ODE you cite is a Cauchy-Euler equation, and we may transform it into an equation with constant coefficients by using the substitution $t=e^x$. We are given to solve:

$\displaystyle 3t^2\frac{d^2y}{dt^2}+6t\frac{dy}{dt}=1$

From the suggested substitution, it follows from the chain rule that:

$\displaystyle \frac{dy}{dx}=\frac{dy}{dt}\frac{dt}{dx}=\frac{dy}{dt}e^x=t\frac{dy}{dt}$

and hence:

(1) $\displaystyle t\frac{dy}{dt}=\frac{dy}{dx}$

Differentiating (1) with respect to $x$, we find from the product rule that:

$\displaystyle \frac{d^2y}{dx^2}=\frac{d}{dx}\left(t\frac{dy}{dt} \right)=\frac{dt}{dx}\frac{dy}{dt}+t\frac{d}{dx} \left(\frac{dy}{dt} \right)$

$\displaystyle \frac{d^2y}{dx^2}=\frac{dy}{dx}+t\frac{d^2y}{dt^2}\frac{dt}{dx}=\frac{dy}{dx}+t\frac{d^2y}{dt^2}e^x$

$\displaystyle \frac{d^2y}{dx^2}=\frac{dy}{dx}+t^2\frac{d^2y}{dt^2}$

and hence:

(2) $\displaystyle t^2\frac{d^2y}{dt^2}=\frac{d^2y}{dx^2}-\frac{dy}{dx}$

Substituting into the given ODE the expressions given in (1) and (2), we obtain:

$\displaystyle 3\left(\frac{d^2y}{dx^2}-\frac{dy}{dx} \right)+6\left(\frac{dy}{dx} \right)=1$

$\displaystyle \frac{d^2y}{dx^2}+\frac{dy}{dx}=\frac{1}{3}$

The characteristic roots are $r=-1,\,0$ and so the solution to the corresponding homogeneous equation is:

$y_h(x)=c_1+c_2e^{-x}$

To ensure linear independence, we must then assume a particular solution of the form:

$y_p(x)=Ax$

And so we compute:

$y_p'(x)=A$

$y_p''(x)=0$

and by substitution, we find:

$\displaystyle 0+A=\frac{1}{3}$ and thus we have:

$\displaystyle y_p(x)=\frac{1}{3}x$ and by superposition, we find:

$\displaystyle y(x)=y_h(x)+y_p(x)=c_1+c_2e^{-x}+\frac{1}{3}x$

Back substituting for $t$, we obtain:

$\displaystyle y(t)=c_1+\frac{c_2}{t}+\frac{1}{3}\ln(t)$

Method 2:

We are given to solve:

$\displaystyle 3t^2\frac{d^2y}{dt^2}+6t\frac{dy}{dt}=1$

Let:

$\displaystyle u(t)=\frac{dy}{dt}\,\therefore\,\frac{du}{dt}= \frac{d^2y}{dt^2}$

and we have:

$\displaystyle 3t^2\frac{du}{dt}+6t\cdot u=1$

$\displaystyle t^2\frac{du}{dt}+2t\cdot u=\frac{1}{3}$

Observe that we may write the left side of the equation as the differentiation of a product:

$\displaystyle \frac{d}{dt}\left(t^2u \right)=\frac{1}{3}$

Integrating with respect to $t$, we obtain:

$\displaystyle t^2u=\frac{1}{3}t+c_1$

Divide through by $t^2$:

$\displaystyle u=\frac{1}{3t}+\frac{c_1}{t^2}$

Back substitute for $u$:

$\displaystyle \frac{dy}{dt}=\frac{1}{3t}+\frac{c_1}{t^2}$

Integrate with respect to $t$:

$\displaystyle y(t)=\frac{1}{3}\ln(t)+\frac{c_1}{t}+c_2$

And so we may conclude:

$\displaystyle f(t)=\frac{1}{3}\ln(t)$

$\displaystyle g(t)=\frac{1}{t}$

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#### Fernando Revilla

##### Well-known member
MHB Math Helper
Another way: using the substitution $v=y'$ we get: $v'+\dfrac{2}{t}v=\dfrac{1}{3t^2}$ (linear equation). According to a well-known theorem its general solution is

$ve^{\int\frac{2}{t}\;dt}-\displaystyle\int \dfrac{1}{3t^2}e^{\int\frac{2}{t}\;dt}\;dt=C$

We get $vt^2-\dfrac{1}{3}t=C$, hence $y'=\dfrac{C}{t^2}+\dfrac{1}{3t}$. Integrating:

$y=\dfrac{-C}{t}+\dfrac{1}{3}\ln t+K=\dfrac{1}{3}\ln t+\dfrac{c_1}{t}+c_2$

Then, $f(t)=\dfrac{1}{3}\ln t,\;g(t)=\dfrac{1}{t}$.

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