Given the equation x^2(y'') + ax(y') + 4(y) = 0 find all values of "a" so that all solutions approach zero as x approaches infinity.
I got the answer (-infinity,-4) U (1,5] but the computer says it is wrong. Help is much appreciated.
I got the answer (-infinity,-4) U (1,5] but the computer says it is wrong. Help is much appreciated.
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x² y'' + ax y' + 4y = 0
y'' + a/x y' + 4/x² y = 0
Using substitution: x = e^t, we get:
d²y/dt² + (a−1) dy/dt + 4 y = 0
We now have second order homogeneous linear equation with constant coefficients. To solve, we find roots of characteristic equation:
r² + (a−1) r + 4 = 0
r = (1 − a ± √(a²−2a−15)) / 2
r₁ = (1 − a + √(a²−2a−15)) / 2
r₂ = (1 − a − √(a²−2a−15)) / 2
————————————————————
Case 1: roots are real
a² − 2a − 15 ≥ 0
(a + 3) (a − 5) ≥ 0
a ≤ −3 or a ≥ 5
y₁ = c₁ e^(r₁t) = c₁ e^(r₁ ln x) = c₁ e^(ln(x^r₁)) = c₁ x^r₁
y₂ = c₂ e^(r₂t) = c₂ e^(r₂ ln x) = c₂ e^(ln(x^r₂)) = c₂ x^r₂
y = c₁ x^r₁ + c₂ x^r₂
If lim[x→∞] (c₁ x^r₁ + c₂ x^r₂) = 0 for all solutions, then we need:
lim[x→∞] (c₁ x^r₁) = 0 and lim[x→∞] (c₂ x^r₂) = 0
i.e. r₁ < 0 and r₂ < 0
(1 − a ± √(a²−2a−15)) / 2 < 0
1 − a ± √(a²−2a−15) < 0
When a ≤ −3
−a ≥ 3
1 − a ≥ 4
1 − a + √(a²−2a−15) ≥ 4
One of the roots > 0
Therefore, when roots are real, then there are no values of a such that
lim[x→∞] (c₁ x^r₁ + c₂ x^r₂) = 0 for all values of c₁ and c₂
————————————————————
Case 2: roots are complex
a² − 2a − 15 ≥ 0
(a + 3) (a − 5) ≥ 0
−3 ≤ a ≤ 5
r = α ± βi = (1−a)/2 + βi
y₁ = c₁ e^(αt) sin(βt) = c₁ e^(α ln x) sin(β ln x) = c₁ x^α sin(β ln x)
y₂ = c₂ e^(αt) cos(βt) = c₂ e^(α ln x) cos(β ln x) = c₂ x^α cos(β ln x)
y = c₁ x^α sin(β lnx) + c₂ x^α cos(β lnx)
If lim[x→∞] (c₁ x^α sin(β lnx) + c₂ x^α cos(β lnx)) = 0 for all solutions,
then we need: l
im[x→∞] (c₁ x^α sin(β lnx)) = 0 and lim[x→∞] (c₂ x^α cos(β lnx)) = 0
Since sin(β lnx) and cos(β lnx) fluctuate between −1 and 1, we need:
lim[x→∞] (c₁ x^α) = 0 and lim[x→∞] (c₂ x^α) = 0 i.e. α < 0
α < 0
(1−a)/2 < 0
1 − a < 0
−a < −1
a > 1
Now since roots are complex, we also have −3 ≤ a ≤ 5
Solution: 1 < a ≤ 5
So it's just (1, 5]
y'' + a/x y' + 4/x² y = 0
Using substitution: x = e^t, we get:
d²y/dt² + (a−1) dy/dt + 4 y = 0
We now have second order homogeneous linear equation with constant coefficients. To solve, we find roots of characteristic equation:
r² + (a−1) r + 4 = 0
r = (1 − a ± √(a²−2a−15)) / 2
r₁ = (1 − a + √(a²−2a−15)) / 2
r₂ = (1 − a − √(a²−2a−15)) / 2
————————————————————
Case 1: roots are real
a² − 2a − 15 ≥ 0
(a + 3) (a − 5) ≥ 0
a ≤ −3 or a ≥ 5
y₁ = c₁ e^(r₁t) = c₁ e^(r₁ ln x) = c₁ e^(ln(x^r₁)) = c₁ x^r₁
y₂ = c₂ e^(r₂t) = c₂ e^(r₂ ln x) = c₂ e^(ln(x^r₂)) = c₂ x^r₂
y = c₁ x^r₁ + c₂ x^r₂
If lim[x→∞] (c₁ x^r₁ + c₂ x^r₂) = 0 for all solutions, then we need:
lim[x→∞] (c₁ x^r₁) = 0 and lim[x→∞] (c₂ x^r₂) = 0
i.e. r₁ < 0 and r₂ < 0
(1 − a ± √(a²−2a−15)) / 2 < 0
1 − a ± √(a²−2a−15) < 0
When a ≤ −3
−a ≥ 3
1 − a ≥ 4
1 − a + √(a²−2a−15) ≥ 4
One of the roots > 0
Therefore, when roots are real, then there are no values of a such that
lim[x→∞] (c₁ x^r₁ + c₂ x^r₂) = 0 for all values of c₁ and c₂
————————————————————
Case 2: roots are complex
a² − 2a − 15 ≥ 0
(a + 3) (a − 5) ≥ 0
−3 ≤ a ≤ 5
r = α ± βi = (1−a)/2 + βi
y₁ = c₁ e^(αt) sin(βt) = c₁ e^(α ln x) sin(β ln x) = c₁ x^α sin(β ln x)
y₂ = c₂ e^(αt) cos(βt) = c₂ e^(α ln x) cos(β ln x) = c₂ x^α cos(β ln x)
y = c₁ x^α sin(β lnx) + c₂ x^α cos(β lnx)
If lim[x→∞] (c₁ x^α sin(β lnx) + c₂ x^α cos(β lnx)) = 0 for all solutions,
then we need: l
im[x→∞] (c₁ x^α sin(β lnx)) = 0 and lim[x→∞] (c₂ x^α cos(β lnx)) = 0
Since sin(β lnx) and cos(β lnx) fluctuate between −1 and 1, we need:
lim[x→∞] (c₁ x^α) = 0 and lim[x→∞] (c₂ x^α) = 0 i.e. α < 0
α < 0
(1−a)/2 < 0
1 − a < 0
−a < −1
a > 1
Now since roots are complex, we also have −3 ≤ a ≤ 5
Solution: 1 < a ≤ 5
So it's just (1, 5]