What is the general solution of x²y'' - 2xy' + 2y = 0 for x > 0?
Ay = c₁eˣ + c₂e²ˣ
By = c₁x + c₂x²
Cy = c₁x + c₂x ln x
Dy = c₁x⁻¹ + c₂x⁻²
Substituting y = xʳ gives r(r - 1) - 2r + 2 = r² - 3r + 2 = 0, so r = 1 and r = 2. The general solution is y = c₁x + c₂x². The exponential answer comes from wrongly treating the equation as constant-coefficient.
Question 2 Short Answer
For x²y'' + 3xy' + y = 0, the indicial equation has a repeated root. Find the general solution for x > 0.
Think about your answer, then reveal below.
Model answer: y = (c₁ + c₂ ln x)/x
The indicial equation is r(r - 1) + 3r + 1 = r² + 2r + 1 = (r + 1)² = 0, so r = -1 is repeated. The solutions are x⁻¹ and x⁻¹ ln x.
Question 3 True / False
The equation x²y'' + xy' + y = 0 has solutions cos(ln x) and sin(ln x) for x > 0.
TTrue
FFalse
Answer: True
The indicial equation is r(r - 1) + r + 1 = r² + 1 = 0, so r = ±i. Complex roots α ± βi give x^α cos(β ln x) and x^α sin(β ln x); here α = 0 and β = 1.
Question 4 Short Answer
Why does the substitution x = eᵗ turn a Cauchy-Euler equation into a constant-coefficient equation?
Think about your answer, then reveal below.
Model answer: With x = eᵗ, x·dy/dx = dy/dt and x²·d²y/dx² = d²y/dt² - dy/dt, so every term loses its power of x and the coefficients become constants.
The new equation is a·y'' + (b - a)y' + cy = 0 in t, whose characteristic equation ar² + (b - a)r + c = 0 is the same as the indicial equation. The solution eʳᵗ becomes xʳ and teʳᵗ becomes xʳ ln x.