Solving Nonhomogeneous Euler Differential Equations Using Substitution
Definition
The differential equation given as follows is called the Euler differential equation.
$$ \begin{align} && a_2 x^2 \dfrac{d^2 y}{d x^2} + a_{1} x \dfrac{dy}{dx} + a_{0} y &= f(x) \label{eq1} \\ \mathrm{or}&& a_2 x^2 y^{\prime \prime} + a_{1} x y^{\prime} +a_{0} y &= f(x) \nonumber \\ \mathrm{or}&& x^2 y^{\prime \prime} + \alpha x y^{\prime} + \beta y &= f(x) \nonumber \end{align} $$
Explanation
It is also referred to as the Euler-Cauchy equation.
If it is a homogeneous equation like $f(x)=0$, it can be relatively easily solved.
If it is in the form of a non-homogeneous equation as above, it is not easy to solve because the coefficients contain the independent variable. It is much more difficult compared to differential equations with constant coefficients. This article introduces a method of solving Euler equations easily through substitution among several methods. The key is substituting $x$ with $e^z$. If we set it as $x=e^z$, the Euler equation becomes a second-order linear differential equation with constant coefficients. Then, it can be easily solved by referring to solutions for second-order non-homogeneous differential equations.
Solution
Step 0.
Substitute with $x=e^z$. Then
$$ \ln x = z, \quad \quad \dfrac{1}{x}=\dfrac{dz}{dx} $$
Step 1.
$$ \begin{align*} x \dfrac{dy}{dx} &= x\dfrac{dy}{dz} \dfrac{dz}{dx} \\ &= x \dfrac{dy}{dz} \dfrac{1}{x} \\ &= \dfrac{dy}{dz} \end{align*} $$
$$ \implies x\dfrac{dy}{dx} = \dfrac{dy}{dz} $$
The first equality is established by the chain rule.
Step 2.
$$ \begin{align*} x^2 \dfrac{ d^2 y}{d x^2} &=x^2 \dfrac{d}{dx} \left( \dfrac{dy}{dx} \right) \\ &= x^2 \dfrac{d}{dx} \left( \dfrac{1}{x} \dfrac{dy}{dz} \right) \\ &= -\dfrac{dy}{dz} + x\dfrac{d}{dx} \left( \dfrac{dy}{dz} \right) \\ &= -\dfrac{dy}{dz} + x\dfrac{d}{dz} \left( \dfrac{dy}{dz} \right) \dfrac{dz}{dx} \\ &= -\dfrac{dy}{dz} + x \dfrac{d^2y}{dz^2} \dfrac{1}{x} \end{align*} $$
$$ \implies x^2 \dfrac{ d^2 y}{d x^2} = \dfrac{d^2y}{dz^2} -\dfrac{dy}{dz} $$
The second equality is established by the result of Step 1. The fourth equality is established by the chain rule. The last equality is established by the result of Step 0.
Step 3.
By substituting the results of Step 1-2. into the Euler equation $\eqref{eq1}$,
$$ a_2 \left( \dfrac{d^2y}{dz^2} -\dfrac{dy}{dz} \right) + a_{1} \dfrac{dy}{dz} + a_{0} y =f(e^z) $$
$$ \implies a_2 \dfrac{d^2y}{dz^2} + (a_{1}-a_2) \dfrac{dy}{dz} + a_{0} y =f(e^z) $$
Now, it can be solved according to the solution methods for second-order non-homogeneous differential equations.
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