Cauchy-Euler Equation with Solved Problems

Cauchy-Euler equation (or Cauchy’s linear equation) is a differential equation with non-constant coefficients i.e, having variable-coefficients. It has the following form

where a₀, a₁, …, a_n are real constants, and f(x) is a function of x.

How to Solve Cauchy-Euler Equation?

The below steps to be followed in order to solve a Cauchy-Euler equation.

1. Make the given differential equation to a differential equation with constant coefficients.
2. For step 1, we usually need the substitution $x=e^z$.
3. Then solve the differential equation with constant coefficients by finding its complimentary function and particular integral.

Solved Problems

Question 1: Solve the following Cauchy-Euler equation $x^2\dfrac{d^2y}{dx^2}-2x\dfrac{dy}{dx}+2y=x^3$.

Solution:

The given equation is $x^2\dfrac{d^2y}{dx^2}-2x\dfrac{dy}{dx}+2y=x^3$ …(I)

Put x=e^z.

∴ $\dfrac{dx}{dz}=e^z$.

Now,

$\dfrac{dy}{dx}=\dfrac{dy}{dz} \cdot \dfrac{dz}{dz}=\dfrac{dy}{dz}e^{-z}$ ⇒ $\dfrac{dy}{dx}=e^{-z}\dfrac{dy}{dz}$ …(II)

Therefore, $\dfrac{d^2y}{dx^2}=$ $\dfrac{d}{dx} \left(e^{-z}\dfrac{dy}{dz} \right)$

So, $\dfrac{d^2y}{dx^2}=$ $\dfrac{d}{dz} \left(e^{-z}\dfrac{dy}{dz} \right) \cdot \dfrac{dz}{dx}$

⇒ $\dfrac{d^2y}{dx^2}=$ $\left(e^{-z}\dfrac{d^2y}{dz^2}-e^{-z}\dfrac{dy}{dz} \right) \cdot e^{-z}$, by product rule of derivatives.

⇒ $\dfrac{d^2y}{dx^2}=e^{-2z}\left(\dfrac{d^2y}{dz^2}-\dfrac{dy}{dz} \right)$ …(III)

Now, using (III) and (II) in the given equation (I), we obtain that $x^2e^{-2z}\left(\dfrac{d^2y}{dz^2}-\dfrac{dy}{dz} \right)$ $-2xe^{-z}\dfrac{dy}{dz}+2y=e^{3z}$ as x=ez. Using x=ez the above equation reduces to $\left(\dfrac{d^2y}{dz^2}-\dfrac{dy}{dz} \right)-2\dfrac{dy}{dz}+2y=e^{3z}$

Putting $D^n\equiv \dfrac{d^n}{dz^n}$, we have:

$(D^2-3D+2)y=e^{3z}$ …(IV)

$\boxed{\text{CF :}}$ The auxiliary equation is given by m²-3m+2 =0

⇒ (m-1)(m-2) = 0

⇒ m=1, 2 (real and unequal roots).

So the complementary function of (IV) is C₁e^x+C₂e^2x where C₁, C₂ are arbitrary constants.

$\boxed{\text{PI :}}$ The particular integral is given by

$\dfrac{1}{D^2-3D+2}e^{3z}$

= $\dfrac{1}{(D-1)(D-2)}e^{3z}$

= $\dfrac{e^{3z}}{(3-1)(3-2)}$

= $\dfrac{e^{3z}}{2}$

Therefore, the complete solution of the given equation is as follows: y = CF + PI = C₁e^x+C₂e^2x + $\dfrac{e^{3z}}{2}$ = C₁lnx+C₂(lnx)² + x²/2 where C₁ and C₂ are arbitrary constants.

Question 2: Solve the following Cauchy-Euler equation $x^4\dfrac{d^3y}{dx^3}+2x^3\dfrac{d^2y}{dx^2}-x^2\dfrac{dy}{dx}+xy=1$.

Solution:

Related Article: Variation of Parameters with Solved Examples