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Initial value problem y''−4y'+3y

Short answer

Auxiliary roots are 1 and 3, so y_c = C_1 e^x + C_2 e^{3x}. Trial y_p = A e^{-2x} gives A = 2/3. Applying y(0)=1 and y'(0)=3 yields C_1 = -5/3, C_2 = 2. Final answer: y = -(5/3)e^x + 2e^{3x} + (2/3)e^{-2x}.

Solve the initial value problem: y'' − 4y' + 3y = 10e⁻²ˣ, y(0) = 1, y'(0) = 3.

Equation: y″ − 4y′ + 3y = 10e^(−2x), y(0) = 1, y′(0) = 3. A linear equation with constant coefficients: y = y_c + y_p.

Complementary function

Auxiliary equation m² − 4m + 3 = 0 gives (m − 1)(m − 3) = 0, so m = 1, 3.

y_c = C₁eˣ + C₂e^(3x).

Particular integral

Try y_p = Ae^(−2x). Then y_p′ = −2Ae^(−2x), y_p″ = 4Ae^(−2x):

4A + 8A + 3A = 10   →   15A = 10   →   A = 2/3

(Equivalently y_p = 10e^(−2x)/f(−2), with f(D) = D² − 4D + 3, f(−2) = 4 + 8 + 3 = 15.)

General solution:

y = C₁eˣ + C₂e^(3x) + (2/3)e^(−2x)

Apply the initial conditions

y′ = C₁eˣ + 3C₂e^(3x) − (4/3)e^(−2x).

y(0) = 1:   C₁ + C₂ + 2/3 = 1         →  C₁ + C₂ = 1/3
y′(0) = 3:  C₁ + 3C₂ − 4/3 = 3        →  C₁ + 3C₂ = 13/3

Subtract: 2C₂ = 4, so C₂ = 2, and C₁ = 1/3 − 2 = −5/3.

Answer: y = −(5/3)eˣ + 2e^(3x) + (2/3)e^(−2x).

Check: y(0) = −5/3 + 2 + 2/3 = 1 ✓; y′(0) = −5/3 + 6 − 4/3 = 3 ✓.

✓ Answer checked by Claude Opus 5.5 (high effort)

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