如何将耦合一阶微分方程组转化为关于y(t)的二阶常微分方程？

Got it, let's work through this step by step—you were on the right track with taking derivatives, but the key was missing one critical step: eliminating (x) by expressing it in terms of (y) and its derivatives, instead of dividing the second-order equations. Here's how to do it properly:

Step 1: Lay out the original coupled equations #

First, let's write your two first-order ODEs clearly:

• (\frac{dy}{dt} = -y + 3x) (we'll call this Equation 1)
• (\frac{dx}{dt} = 4x - 2y) (this is Equation 2)

Step 2: Differentiate (y'(t)) to get (y''(t)) #

Take the derivative of both sides of Equation 1 with respect to (t):
(\frac{d^2y}{dt2} = -\frac{dy}{dt} + 3\frac{dx}{dt})

Step 3: Substitute (y'(t)) and (x'(t)) using the original equations #

We already know (y'(t)) from Equation 1, and (x'(t)) from Equation 2. Plug those into the second-derivative equation:
(y'' = -(-y + 3x) + 3(4x - 2y))

Now expand and simplify this:
(y'' = y - 3x + 12x - 6y)
(y'' = -5y + 9x)
(Quick note: you had a sign error in your initial calculation here—this is the correct simplified form)

Step 4: Solve Equation 1 for (x) in terms of (y) and (y') #

This is the step you skipped earlier! From Equation 1, rearrange to isolate (x):
(3x = y' + y)
(x = \frac{y' + y}{3})

Step 5: Substitute (x) into the (y'') equation #

Now replace (x) in the (y'' = -5y +9x) expression with the formula we just found:
(y'' = -5y + 9\left(\frac{y' + y}{3}\right))

Simplify the right-hand side:
(y'' = -5y + 3(y' + y))
(y'' = -5y + 3y' + 3y)
(y'' = 3y' - 2y)

Final Result #

Rearrange terms to get the standard form of a second-order linear ODE for (y(t)):
(\frac{d^2y}{dt2} - 3\frac{dy}{dt} + 2y = 0)

The core idea here is that for coupled first-order systems, you eliminate the extra variable (in this case, (x)) by expressing it using the target variable ((y)) and its derivatives, then substitute back to get a single equation only in (y(t)) and its derivatives.

内容的提问来源于stack exchange，提问作者oldselflearner1959