Consider the system of ODEs $$\dfrac{dx}{dt} = x + y, \qquad \dfrac{dy}{dt} = y$$ The general solution of this system of equations is

Question

BlackTom AI · Accepted Answer

Since dy/dt = y, integrate to get y(t) = B e^{t} where B is an arbitrary constant.

1) Relevant concepts
- Solving decoupled linear ODEs: dy/dt = y is separable/exponential.
- First-order linear ODE for x with an inhomogeneous term y(t).
- Matrix viewpoint: A = [[1,1],[0,1]] is a Jordan block with eigenvalue 1 (double), so solutions include t e^{t} terms.

2) Complete reasoning step by step
From dy/dt = y:
- dy/y = dt ⇒ ln|y| = t + C1 ⇒ y = B e^{t} (B arbitrary).

Now substitute y into the x-equation:
dx/dt = x + y = x + B e^{t}.
Rewrite as a linear ODE in standard form:
dx/dt - x = B e^{t}.

Use integrating factor μ(t) = e^{-t}:
d/dt[x e^{-t}] = B e^{t} e^{-t} = B.

Integrate both sides:
x e^{-t} = ∫ B dt = B t + C, where C is an arbitrary constant.

Solve for x:
x(t) = (B t + C) e^{t} = B t e^{t} + C e^{t}.

Rename constants if desired: let A = C, keep B as the constant from y. So
x(t) = (A + B t) e^{t},  y(t) = B e^{t}.

3) Intermediate steps (displayed)
- y = B e^{t}
- dx/dt - x = B e^{t}
- μ = e^{-t}
- d/dt[x e^{-t}] = B ⇒ x e^{-t} = B t + C ⇒ x = (B t + C) e^{t}

4) Why this is correct
The solution satisfies both equations for arbitrary constants A, B. The presence of the Bt e^{t} term is necessary because the inhomogeneous term y = B e^{t} resonates with the homogeneous solution e^{t}, producing a secular t e^{t} term (consistent with Jordan-block behavior).

5) Why the other choices are incorrect
- (a) x = A e^{t}, y = B e^{t} omits the required B t e^{t} term, so it does not satisfy dx/dt = x + y unless B = 0.
- (b) x = 0, y = B e^{t} fails unless both constants satisfy special initial conditions; x ≡ 0 does not satisfy dx/dt = x + y when y ≠ 0.
- (c) x = A (constant), y = B e^{t} contradicts dx/dt = x + y unless A and B are specially constrained (not general).

6) Conclusion
The correct general solution is x(t) = (A + B t) e^{t}, y(t) = B e^{t}, so none of the listed options (a)–(c) match; therefore answer d ("None of the answers above") is correct.

Consider the system of ODEs \[\dfrac{dx}{dt} = x + y, \qquad \dfrac{dy}{dt} = y\] The general solution of this system of equations is单项选择题

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