A system with variable coefficients. We find a homogeneous solution using a power-of-t ansatz,
then use reduction of order (change of variables) to find the general solution of the non-homogeneous system.
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Step 1: Homogeneous Solution - Power Ansatz
Find a solution of the homogeneous system whose components are powers of t
Guiding Question
Try a solution of the form x(t) = atα, y(t) = btβ
Substitute into the system and find α, β and the ratio a:b
Hint: from the equation structure α = β + 2. Try β = 3
Step c: Substitution into the system and simplification
After substitution and cancellation of the homogeneous terms (which cancel thanks to the homogeneous solution),
two separate scalar equations are obtained
The Scalar Equations
dξ/dt = (5/t)ξ − 3t3 dη/dt = t−6ξ − 4t−2
Key insight:
The system "decouples" - first solve the ξ equation (alone), then substitute ξ into the η equation.
Step 3 Solution - The ξ Equation
The equation:
dξ/dt − (5/t)ξ = −3t3
Integrating factor:
μ = e−∫(5/t)dt = e−5 ln t = t−5
Multiply by integrating factor:
d/dt(t−5ξ) = −3t−2
Integration:
t−5ξ = −3 · ∫t−2dt = 3t−1 + A
ξ = 3t4 + At5
The Solution
ξ = 3t4 + At5
Step 4 Solution - The η Equation
Substituting ξ into the equation:
dη/dt = t−6(3t4 + At5) − 4t−2
Simplification:
dη/dt = 3t−2 + At−1 − 4t−2
dη/dt = −t−2 + At−1
Term-by-term integration:
η = ∫(−t−2)dt + A·∫(t−1)dt + B
η = t−1 + A·ln(t) + B
Note:
A is the same constant from the ξ solution, and B is a new integration constant.
The Solution
η = t−1 + A·ln(t) + B
Step 5 Solution - The General Solution
Back-substitution for x:
x = ξ − 2t5η
= (3t4 + At5) − 2t5(t−1 + A·ln t + B)
= 3t4 + At5 − 2t4 − 2At5·ln t − 2Bt5
= t4 + A(t5 − 2t5·ln t) − 2Bt5
Back-substitution for y:
y = t3η = t3(t−1 + A·ln t + B)
= t2 + At3·ln t + Bt3