Answer to Question #325046 in Differential Equations for Thami

Question

(D+2)x-3y=1

-3x+(D+2)y=e^-t

Answer 1

Multiplying first equation by 3 and solving second for 3x we’ll get

(D+2)3x-9y=3

3x = (D+2)y - e^-t  

Substituting 3x from second equation into first

(D+2)[(D+2)y - e^-t ] – 9y = 3

(D+2)²y + e^-t – 2e^-t  – 9y = 3

D²y+4Dy+4y – 9y = 3+e^-t  

D²y + 4Dy – 5y = 3 +e^-t  

Characteristic equation

λ² + 4λ – 5 = 0  =>   $\lambda_{1,2}=-2\pm\sqrt9$    => λ₁ = 1, λ₂ = -5

So the solution of homogeneous equation is y₀(t) = C₁e^t+C₂e^-5t, where C₁, C₂ are arbitrary constants.

Partial solution may be find in the form y₁(t) = A+Be^-t. Substitution this into equation gives

(Be^-t – 4Be^-t – 5A – 5Be^-t) = 3 +e^-t  => -5A = 3; -8B = 1 => A = -0.6; B = -0.125

So y(t) = y₀(t) + y₁(t) = C₁e^t+C₂e^-5t – 0.6 – 0.125e^-t 

From second equation of the system

x = (1/3)[ (D+2)y - e^-t] = (1/3)[ C₁e^t – 5C₂e^-5t + 2C₁e^t+2C₂e^-5t +0.125e^-t – 1.2 – 0.25 e^-t – e^-t] = (1/3)[ 3C₁e^t – 3C₂e^-5t – 1.2 – 1.125 e^-t ] = C₁e^t – C₂e^-5t – 0.4 – 0.375 e^-t

Solution

x(t) = C₁e^t – C₂e^-5t – 0.4 – 0.375 e^-t

y(t) = C₁e^t+C₂e^-5t – 0.6 – 0.125e^-t