Chapter 2 · Second-Order ODEs

Euler Equations & Reduction of Order

Solving Variable-Coefficient Equations

Dr. Mohamed Mabrok · Qatar University

The Big Picture: Second-Order Linear ODEs

Constant
Coefficients

Homogeneous

Non-Homog.

ay'' + by' + cy = g(x)

Variable
Coefficients

Euler Equations ✓

Reduction of Order ✓

What Makes Euler Equations Special?

Standard Form

      $$ax^2 y'' + bxy' + cy = 0$$
    
      Each coefficient has $x$ raised to the same power as the derivative order!

Constant Coefficients

$$ay'' + by' + cy = 0$$

Try: $y = e^{rx}$

Euler Equations

$$ax^2y'' + bxy' + cy = 0$$

Try: $y = x^r$

The Indicial Equation Method

STEP 1 Assume $y = x^r$ and compute derivatives

y' = rx^{r-1}, \quad y'' = r(r-1)x^{r-2}

STEP 2 Substitute into $ax^2y'' + bxy' + cy = 0$

$$ar(r-1)x^r + brx^r + cx^r = 0$$

STEP 3 Factor out $x^r$ and set bracket to zero

$$ar^2 + (b-a)r + c = 0$$

The Indicial Equation!

      This quadratic in $r$ has the same form as the characteristic equation for constant coefficients!

Three Cases (Parallel to Constant Coefficients)

Case 1: Distinct Real Roots

$r_1 \neq r_2$

$$y = C_1 x^{r_1} + C_2 x^{r_2}$$

Case 2: Repeated Root

$r_1 = r_2 = r$

$$y = x^r(C_1 + C_2 \ln x)$$

Case 3: Complex Roots

$r = \lambda \pm \omega i$

$$y = x^\lambda[C_1\cos(\omega\ln x) + C_2\sin(\omega\ln x)]$$

Worked Example

Solve: $x^2 y'' - xy' - 8y = 0$

STEP 1 Write indicial equation

r(r-1) - r - 8 = 0 \quad \Rightarrow \quad r^2 - 2r - 8 = 0

STEP 2 Factor

(r-4)(r+2) = 0 \quad \Rightarrow \quad r_1 = 4, \quad r_2 = -2

STEP 3 Write general solution

y = C_1 x^4 + \frac{C_2}{x^2}

Worked Example

Solve: $x^2 y'' - 5xy' + 9y = 0$

STEP 1 Write indicial equation

r(r-1) - 5r + 9 = 0 \quad \Rightarrow \quad r^2 - 6r + 9 = 0

STEP 2 Factor as perfect square

(r-3)^2 = 0 \quad \Rightarrow \quad r = 3 \text{ (repeated)}

STEP 3 Use repeated root formula

y = x^3(C_1 + C_2 \ln x)

Worked Example

Solve: $x^2 y'' + 3xy' + 2y = 0$

STEP 1 Write indicial equation

r(r-1) + 3r + 2 = 0 \quad \Rightarrow \quad r^2 + 2r + 2 = 0

STEP 2 Use quadratic formula

r = \frac{-2 \pm \sqrt{4-8}}{2} = \frac{-2 \pm 2i}{2} = -1 \pm i

STEP 3 Identify $\lambda$ and $\omega$

\lambda = -1, \quad \omega = 1

STEP 4 General solution

y = \frac{1}{x}[C_1\cos(\ln x) + C_2\sin(\ln x)]

The Transformation Method

Key Insight

      Let $x = e^t$ (so $t = \ln x$) and define $Y(t) = y(e^t)$. Then the Euler equation transforms into a constant coefficient equation!

EULER $ax^2y'' + bxy' + cy = 0$

↓ Transform: $x = e^t$ ↓

CONST $aY'' + (b-a)Y' + cY = 0$

Why This Works

      This explains the parallel: the transformation converts power functions $x^r$ into exponentials $e^{rt}$, revealing the deep connection between the two equation types!

Reduction of Order: When You Know One Solution

The Setup

      Given: A second-order linear ODE $P_0(x)y'' + P_1(x)y' + P_2(x)y = 0$
    
      And: You know one solution $y_1(x)$

The Brilliant Idea

      $$y = u(x) \cdot y_1(x)$$
    
      Assume the second solution has this form. Substituting and simplifying reduces the problem to a first-order equation in $z = u'$!

The Power

      First-order equations are much easier to solve than second-order. Once you solve for $u'$, integrate to find $u$, and you get the second linearly independent solution!

Reduction of Order: The Derivation

STEP 1 Set $y = uy_1$ and compute derivatives

y' = u'y_1 + uy_1', \quad y'' = u''y_1 + 2u'y_1' + uy_1''

STEP 2 Substitute into ODE and rearrange

$$u[P_0y_1'' + P_1y_1' + P_2y_1] + P_0y_1u'' + (2P_0y_1' + P_1y_1)u' = 0$$

STEP 3 Use fact that $y_1$ solves the ODE

$$P_0y_1u'' + (2P_0y_1' + P_1y_1)u' = 0$$

STEP 4 Let $z = u'$: first-order equation in $z$

$$P_0y_1z' + (2P_0y_1' + P_1y_1)z = 0$$

Worked Example

Solve: $x^2 y'' - 3xy' + 3y = 0$, given $y_1 = x$

STEP 1 Verify $y_1 = x$ solves the ODE

$$y_1' = 1, \quad y_1'' = 0 \quad \Rightarrow \quad x^2(0) - 3x(1) + 3x = 0$$ ✓

STEP 2 Set $y = ux$ and find derivatives

y' = u'x + u, \quad y'' = u''x + 2u'

STEP 3 Substitute and simplify

$$x^2(u''x + 2u') - 3x(u'x + u) + 3ux = 0$$ $$x^3u'' - x^2u' = 0$$

STEP 4 Solve for $u'$: $u' = Cx$

STEP 5 Integrate: $u = \frac{Cx^2}{2} + D$

STEP 6 Second solution: $y_2 = ux = x^3$

$$y = C_1 x + C_2 x^3$$

Non-Homogeneous Example

Non-Homogeneous Euler: $x^2 y'' - xy' + y = x^3$

STEP 1 Solve homogeneous: $x^2 y'' - xy' + y = 0$

r^2 - 2r + 1 = 0 \quad \Rightarrow \quad r = 1 \text{ (repeated)}$$ $$y_c = x(C_1 + C_2\ln x)

STEP 2 Find particular solution (variation of parameters)

y_p = \frac{x^3}{4}

STEP 3 General solution

y = x(C_1 + C_2\ln x) + \frac{x^3}{4}

IVP Example

Solve: $x^2 y'' + xy' - y = 0$, $y(1) = 2$, $y'(1) = -3$

STEP 1 Find general solution

r^2 - 1 = 0 \quad \Rightarrow \quad r = \pm 1$$ $$y = C_1 x + \frac{C_2}{x}

STEP 2 Find derivative

y' = C_1 - \frac{C_2}{x^2}

STEP 3 Apply initial conditions

y(1) = C_1 + C_2 = 2, \quad y'(1) = C_1 - C_2 = -3

STEP 4 Solve for constants

C_1 = -\frac{1}{2}, \quad C_2 = \frac{5}{2}

y = -\frac{x}{2} + \frac{5}{2x}

Key Takeaways & Summary

Euler: Case 1

Distinct real roots $r_1, r_2$

$$y = C_1 x^{r_1} + C_2 x^{r_2}$$

Euler: Case 2

Repeated root $r$

$$y = x^r(C_1 + C_2\ln x)$$

Euler: Case 3

Complex roots $\lambda \pm \omega i$

$$y = x^\lambda[C_1\cos(\omega\ln x) + C_2\sin(\omega\ln x)]$$

Reduction of Order

      If you know one solution $y_1$, assume $y = uy_1$. Substitution gives a first-order equation in $z = u'$. Solve for $z$, integrate to find $u$, and you have the second solution $y_2 = uy_1$.

The Transformation Connection

      The substitution $x = e^t$ transforms Euler equations into constant coefficient equations in $Y(t)$, revealing the deep structural relationship between variable and constant coefficient ODEs.