A normal constant-coefficient ODE like y ′ ′ + a y ′ + b y = 0 y'' + ay' + by = 0 y ′′ + a y ′ + b y = 0 exponentials e r x e^{rx} e r x multiplicative world : each term has x k x^k x k k k k power x m x^m x m y = x m y = x^m y = x m e r x e^{rx} e r x equidimensional says it all: if x x x x k y ( k ) x^k y^{(k)} x k y ( k ) same dimensions , so the equation doesn't care about the scale of x x x
Definition Cauchy-Euler equation (2nd order)A linear ODE of the form
a x 2 y ′ ′ + b x y ′ + c y = 0 a\,x^2 y'' + b\,x\,y' + c\,y = 0 a x 2 y ′′ + b x y ′ + c y = 0 a , b , c a,b,c a , b , c x x x x 2 y ′ ′ x^2 y'' x 2 y ′′ x 1 y ′ x^1 y' x 1 y ′ x 0 y x^0 y x 0 y n n n ∑ k = 0 n a k x k y ( k ) = 0 \sum_{k=0}^n a_k\, x^k y^{(k)} = 0 ∑ k = 0 n a k x k y ( k ) = 0
The defining feature: degree of x x x . This balance is what makes the substitution y = x m y=x^m y = x m
y = x m y=x^m y = x m x m x^m x m m x m − 1 m x^{m-1} m x m − 1 lowers the power by 1. But the coefficient x k x^k x k k k k raises it back by k k k x k d k d x k x m x^k \frac{d^k}{dx^k}x^m x k d x k d k x m always a constant times x m x^m x m . Every term collapses to ( number ) ⋅ x m (\text{number})\cdot x^m ( number ) ⋅ x m x m x^m x m m m m
Try y = x m y = x^m y = x m y ′ = m x m − 1 , y ′ ′ = m ( m − 1 ) x m − 2 . y' = m x^{m-1}, \qquad y'' = m(m-1)x^{m-2}. y ′ = m x m − 1 , y ′′ = m ( m − 1 ) x m − 2 .
Substitute into a x 2 y ′ ′ + b x y ′ + c y = 0 ax^2y'' + bxy' + cy = 0 a x 2 y ′′ + b x y ′ + cy = 0 a x 2 ⋅ m ( m − 1 ) x m − 2 + b x ⋅ m x m − 1 + c x m = 0 a x^2\cdot m(m-1)x^{m-2} + b x\cdot m x^{m-1} + c\, x^m = 0 a x 2 ⋅ m ( m − 1 ) x m − 2 + b x ⋅ m x m − 1 + c x m = 0
Why this step? Each x x x x 2 ⋅ x m − 2 = x m x^2\cdot x^{m-2}=x^m x 2 ⋅ x m − 2 = x m x ⋅ x m − 1 = x m x\cdot x^{m-1}=x^m x ⋅ x m − 1 = x m [ a m ( m − 1 ) + b m + c ] x m = 0. \big[\,a\,m(m-1) + b\,m + c\,\big]\,x^m = 0. [ a m ( m − 1 ) + b m + c ] x m = 0.
For x ≠ 0 x\neq 0 x = 0 = 0 =0 = 0
Case 1 — distinct real roots m 1 ≠ m 2 m_1\neq m_2 m 1 = m 2
y = C 1 x m 1 + C 2 x m 2 . y = C_1 x^{m_1} + C_2 x^{m_2}. y = C 1 x m 1 + C 2 x m 2 .
Case 2 — repeated root m 1 = m 2 = m m_1=m_2=m m 1 = m 2 = m We only get one solution x m x^m x m
ln x \ln x ln x e r x e^{rx} e r x x e r x x\,e^{rx} x e r x x = e t x=e^t x = e t t t t e m t e^{mt} e m t t e m t t\,e^{mt} t e m t t = ln x t=\ln x t = ln x x m x^m x m = = ( ln x ) x m = = ==(\ln x)\,x^m== == ( ln x ) x m ==
y = ( C 1 + C 2 ln x ) x m . y = (C_1 + C_2 \ln x)\,x^m. y = ( C 1 + C 2 ln x ) x m .
Case 3 — complex roots m = α ± i β m = \alpha \pm i\beta m = α ± i β x α ± i β = x α x ± i β = x α e ± i β ln x x^{\alpha\pm i\beta}=x^\alpha x^{\pm i\beta}=x^\alpha e^{\pm i\beta \ln x} x α ± i β = x α x ± i β = x α e ± i β l n x e i θ = cos θ + i sin θ e^{i\theta}=\cos\theta+i\sin\theta e i θ = cos θ + i sin θ θ = β ln x \theta=\beta\ln x θ = β ln x y = x α [ C 1 cos ( β ln x ) + C 2 sin ( β ln x ) ] . y = x^{\alpha}\big[\,C_1\cos(\beta \ln x) + C_2\sin(\beta \ln x)\,\big]. y = x α [ C 1 cos ( β ln x ) + C 2 sin ( β ln x ) ] .
Intuition WHY this substitutionWe suspect Cauchy-Euler is secretly a constant-coefficient equation in disguise. The disguise is the variable x x x x = e t x=e^t x = e t t = ln x t=\ln x t = ln x x x x t t t
Let t = ln x t=\ln x t = ln x d t d x = 1 x \dfrac{dt}{dx}=\dfrac1x d x d t = x 1 d y d x = 1 x d y d t ⇒ x y ′ = d y d t . \frac{dy}{dx}=\frac{1}{x}\frac{dy}{dt}\ \Rightarrow\ x y' = \frac{dy}{dt}. d x d y = x 1 d t d y ⇒ x y ′ = d t d y . y ′ ′ = 1 x 2 ( d 2 y d t 2 − d y d t ) ⇒ x 2 y ′ ′ = d 2 y d t 2 − d y d t . y'' = \frac{1}{x^2}\Big(\frac{d^2y}{dt^2}-\frac{dy}{dt}\Big)\ \Rightarrow\ x^2 y'' = \frac{d^2y}{dt^2}-\frac{dy}{dt}. y ′′ = x 2 1 ( d t 2 d 2 y − d t d y ) ⇒ x 2 y ′′ = d t 2 d 2 y − d t d y .
Why this step? Substituting these into a x 2 y ′ ′ + b x y ′ + c y = 0 ax^2y''+bxy'+cy=0 a x 2 y ′′ + b x y ′ + cy = 0 x x x a ( y ¨ − y ˙ ) + b y ˙ + c y = 0 ⟹ a y ¨ + ( b − a ) y ˙ + c y = 0 , a\Big(\ddot y - \dot y\Big) + b\,\dot y + c\,y = 0 \;\Longrightarrow\; a\,\ddot y + (b-a)\dot y + c\,y = 0, a ( y ¨ − y ˙ ) + b y ˙ + c y = 0 ⟹ a y ¨ + ( b − a ) y ˙ + c y = 0 , constant-coefficient ODE in t t t a m 2 + ( b − a ) m + c = 0 am^2+(b-a)m+c=0 a m 2 + ( b − a ) m + c = 0 identical to the indicial equation. Solve in t t t t = ln x t=\ln x t = ln x t e m t → ( ln x ) x m t e^{mt}\to (\ln x)x^m t e m t → ( ln x ) x m
Common mistake Forgetting the( b − a ) (b-a) ( b − a ) Wrong instinct: "The characteristic equation is just a m 2 + b m + c = 0 am^2+bm+c=0 a m 2 + bm + c = 0 Why it feels right: It mirrors constant-coefficient ODEs where you read off coefficients directly.
Fix: The x 2 y ′ ′ x^2y'' x 2 y ′′ m ( m − 1 ) = m 2 − m m(m-1)=m^2-m m ( m − 1 ) = m 2 − m extra − a m -am − am a m 2 + ( b − a ) m + c = 0 am^2+(b-a)m+c=0 a m 2 + ( b − a ) m + c = 0 a m ( m − 1 ) + b m + c am(m-1)+bm+c am ( m − 1 ) + bm + c
x e . . . x\,e^{...} x e ... ln x \ln x ln x Wrong: writing the second solution as x ⋅ x m x\cdot x^m x ⋅ x m Why it feels right: In constant-coefficient land the second solution is x e r x x e^{rx} x e r x x × x\times x × Fix: The "extra factor" lives in the t t t t e m t t e^{mt} t e m t t = ln x t=\ln x t = ln x ( ln x ) x m \boxed{(\ln x)x^m} ( ln x ) x m not x ⋅ x m x\cdot x^m x ⋅ x m
Common mistake Ignoring the domainWrong: writing ln x \ln x ln x x x x Fix: Solutions involve ln x \ln x ln x x > 0 x>0 x > 0 x < 0 x<0 x < 0 ∣ x ∣ |x| ∣ x ∣ x m → ∣ x ∣ m x^m\to |x|^m x m → ∣ x ∣ m ln x → ln ∣ x ∣ \ln x\to \ln|x| ln x → ln ∣ x ∣ x = 0 x=0 x = 0 y ′ ′ y'' y ′′
Recall Feynman: explain to a 12-year-oldImagine a special spring puzzle where the rule stays the same no matter how big or small you draw it — it doesn't have a favourite size (that's "equidimensional"). For springs that don't care about size, the natural shapes aren't bendy waves but stretchy power curves like x x x x 2 x^2 x 2 1 x \frac1x x 1 x x x ln x \ln x ln x cos ( ln x ) \cos(\ln x) cos ( ln x ) sin ( ln x ) \sin(\ln x) sin ( ln x )
"Powers for equal-dimensions; the auxiliary equation is m ( m − 1 ) m(m-1) m ( m − 1 ) m 2 m^2 m 2
E quidimensional → E xponent guess x m x^m x m x 2 y ′ ′ → m ( m − 1 ) x^2y''\to m(m-1) x 2 y ′′ → m ( m − 1 ) x y ′ → m xy'\to m x y ′ → m y → 1 y\to 1 y → 1 "two-eleven, one-em, one" .Roots: R eal distinct = two powers, R epeated = sprinkle ln x \ln x ln x R otating (complex) = cos / sin \cos/\sin cos / sin ln x \ln x ln x
What substitution converts a Cauchy-Euler equation to constant-coefficient? x = e t x=e^t x = e t , i.e.
t = ln x t=\ln x t = ln x .
For a x 2 y ′ ′ + b x y ′ + c y = 0 ax^2y''+bxy'+cy=0 a x 2 y ′′ + b x y ′ + cy = 0 a m ( m − 1 ) + b m + c = 0 am(m-1)+bm+c=0 am ( m − 1 ) + bm + c = 0 , i.e.
a m 2 + ( b − a ) m + c = 0 am^2+(b-a)m+c=0 a m 2 + ( b − a ) m + c = 0 .
Why does guessing y = x m y=x^m y = x m x k d k d x k x m x^k\frac{d^k}{dx^k}x^m x k d x k d k x m is always a constant times
x m x^m x m , so every term collapses to (number)
⋅ x m \cdot x^m ⋅ x m .
General solution for distinct real roots m 1 , m 2 m_1,m_2 m 1 , m 2 y = C 1 x m 1 + C 2 x m 2 y=C_1x^{m_1}+C_2x^{m_2} y = C 1 x m 1 + C 2 x m 2 .
General solution for a repeated root m m m y = ( C 1 + C 2 ln x ) x m y=(C_1+C_2\ln x)x^m y = ( C 1 + C 2 ln x ) x m .
General solution for complex roots α ± i β \alpha\pm i\beta α ± i β y = x α [ C 1 cos ( β ln x ) + C 2 sin ( β ln x ) ] y=x^\alpha[C_1\cos(\beta\ln x)+C_2\sin(\beta\ln x)] y = x α [ C 1 cos ( β ln x ) + C 2 sin ( β ln x )] .
Where does the ln x \ln x ln x From the
t e m t te^{mt} t e m t second solution in the
t t t -world;
t = ln x t=\ln x t = ln x .
Why is it called "equidimensional"? Every term
x k y ( k ) x^k y^{(k)} x k y ( k ) has the same physical dimensions, so the equation is scale-invariant.
What is x 2 y ′ ′ x^2y'' x 2 y ′′ t = ln x t=\ln x t = ln x y ¨ − y ˙ \ddot y-\dot y y ¨ − y ˙ (and
x y ′ = y ˙ xy'=\dot y x y ′ = y ˙ ).
Is x = 0 x=0 x = 0 No, it's a singular point — the leading coefficient
a x 2 ax^2 a x 2 vanishes there.
Constant-Coefficient Linear ODEs — Cauchy-Euler becomes one via x = e t x=e^t x = e t Characteristic / Auxiliary Equation — same role, here it's the indicial equation.Euler's Formula — converts x i β x^{i\beta} x i β cos ( ln x ) , sin ( ln x ) \cos(\ln x),\sin(\ln x) cos ( ln x ) , sin ( ln x ) Reduction of Order — alternate route to the ln x \ln x ln x Frobenius Method & Regular Singular Points — Cauchy-Euler is the simplest regular singular point; indicial equation generalizes here.Variation of Parameters — for the non-homogeneous case.
substitute, x powers cancel
repeated root gives t e^mt
degree of x = derivative order
Indicial equation a m^2 + b-a m + c = 0
y = C1 + C2 ln x times x^m
y = x^a cos and sin of b ln x
constant-coefficient ODE in t
Intuition Hinglish mein samjhoDekho, Cauchy-Euler equation hota hai jaisa a x 2 y ′ ′ + b x y ′ + c y = 0 ax^2y''+bxy'+cy=0 a x 2 y ′′ + b x y ′ + cy = 0 k k k x k x^k x k x x x
Constant-coefficient ODE mein hum e r x e^{rx} e r x x x x y = x m y=x^m y = x m x x x a m ( m − 1 ) + b m + c = 0 am(m-1)+bm+c=0 am ( m − 1 ) + bm + c = 0 indicial equation hai. Yaad rakho, x 2 y ′ ′ x^2y'' x 2 y ′′ m ( m − 1 ) m(m-1) m ( m − 1 ) m 2 m^2 m 2
Roots se solution decide hota hai. Do alag real roots → C 1 x m 1 + C 2 x m 2 C_1x^{m_1}+C_2x^{m_2} C 1 x m 1 + C 2 x m 2 ln x \ln x ln x ( C 1 + C 2 ln x ) x m (C_1+C_2\ln x)x^m ( C 1 + C 2 ln x ) x m α ± i β \alpha\pm i\beta α ± i β x α [ cos ( β ln x ) x^\alpha[\cos(\beta\ln x) x α [ cos ( β ln x ) sin ( β ln x ) ] \sin(\beta\ln x)] sin ( β ln x )] ln x \ln x ln x x = e t x=e^t x = e t t t t t e m t te^{mt} t e m t t = ln x t=\ln x t = ln x ln x \ln x ln x x > 0 x>0 x > 0 x = 0 x=0 x = 0
