Euler's Equation - Free MCQ Practice Test with solutions, GATE MA Engineering

Substitute x = e^t, so x² y''² → D(D-1)y and x y' → Dy. The auxiliary equation is:
m(m-1) - 4m + 6 = 0
m² - m - 4m + 6 = 0
m² - 5m + 6 = 0
(m-2)(m-3) = 0.
Roots are real and distinct: m₁ = 2, m₂ = 3.
Solution: y = C₁ x² + C₂ x³.

Substitute x = e^t. The auxiliary equation becomes:
m(m-1) + 3m + 5 = 0
⇒ m² + 2m + 5 = 0.
Roots: .
The solution in terms of t:
y(t) = e^-t [C₁ cos(2t) + C₂ sin(2t)].
Substitute e^-t = x^-1 and t = x: y(x) = x^-1 [C₁ cos(2 ln x) + C₂ sin(2 lnx)].

Transform to constant coefficients with x = e^t:
[D(D-1) - 2D + 2]y = e^3t
⇒ [m² - 3m + 2]y = e^3t.
For P.I., substitute D = 3 (coefficient of t in exponent) into the operator
​
​.
Substitute back e^3t = (e^t)³=x³:
​

Substitute x = e^t, so ln x = t.
[D(D-1) + D - 1]y = 4t
⇒ [D² - 1]y = 4t.
.
Using binomial expansion: (1-z)^-1 = 1 + z + z² .......
P.I. = - [1 + D² + D⁴ + ...] (4t).
Since D²(4t) = 0:
P.I. = - [1](4t) = -4t.
Substitute t = ln x:
P.I. = -4 ln x.

Auxiliary equation:
m(m-1) + 5m + 4 = 0
⇒ m² + 4m + 4 = 0
⇒ (m+2)² = 0.
Roots are real and repeated: m = -2, -2.
Solution for repeated roots:(m,m):
y = x^m (C₁ + C₂ ln x)
y = x^-2 (C₁ + C₂ ln x)

Auxiliary equation:
m(m-1) + m = 0
m² - m + m = 0
m² = 0.
Roots are repeated: m = 0, 0.
Using the repeated root formula y = x^m(C₁ + C₂ ln x) with m=0:
y = x⁰ (C₁ + C₂ ln x)
y = 1.(C₁ + C₂ ln x)
y= C₁ + C₂ ln x.

This is a homogeneous Cauchy-Euler equation. We assume a solution of the form y = x^m.

Substitute into the equation:

The auxiliary equation is:
m² - m - 3m + 3 = 0
m² - 4m + 3 = 0
(m-1)(m-3) = 0
The roots are real and distinct: m₁ = 1, m₂ = 3.
The general solution is:

Substitute x = e^t or t = ln x. Let . The equation transforms to constant coefficients using the operator identities x y' = Dy and x² y'' = D(D-1)y.
[D(D-1) + D + 1]y = 0
[D² - D + D + 1]y = 0
[D² + 1]y = 0
The auxiliary equation is m² + 1 = 0, so roots are m =±i (complex conjugates  α ± iβ
where α=0, β=1.
The solution in terms of  is:
y(t) = e^0t [C₁ cos(1t) + C₂ sin(1t)]
y(t) = C₁ cos t + C₂ sin t
Substitute back t = ln x:
y(x) = C₁ cos(\ln x) + C₂ sin(ln x)

Substitute x = e^t, so t = ln x.
The equation becomes:
[D(D-1) - D + 1]y = 0
[D² - 2D + 1]y = 0
(D-1)² y = 0

The roots are real and repeated: m = 1, 1.
The solution in terms of t is:
y(t) = (C₁ + C₂ t)e^t
Substitute back t = ln x and e^t = x:
y(x) = (C₁ + C₂ ln x)x
y(x) = C₁ x + C₂ x ln x
Now apply boundary conditions:
1. y(1) = 1:
1 = C₁(1) + C₂(1)(0) ⇒ C₁ = 1
2. Differentiate y(x):

Apply y'(1) = 2:
2 = C₁ + C₂(0 + 1)
2 = 1 + C₂ ⇒ C₂ = 1
Substitute constants back:
y = 1x + 1x ln x = x(1 + ln x)

The standard method to linearize a Cauchy-Euler equation (which has variable coefficients xⁿ) into a differential equation with constant coefficients is the substitution:   
x = e^t  or t = ln x   
This substitution converts terms like , allowing the use of standard auxiliary equation methods.