一阶常微分方程通解求解疑问：$(2x-3)\frac{dx}{dt}+t-2=0$

Let's walk through how to get from your integrated quadratic equation to the provided solution form—it's all about completing the square for the (x)-terms and redefining arbitrary constants (a common trick to clean up messy expressions in differential equations!).

Here's the step-by-step breakdown:

1. Start with your integrated result
   You correctly separated variables and integrated to reach:

   x^2 - 3x = -\frac{t^2}{2} + 2t + C
   

2. Complete the square for the (x)-terms
   Rewrite the left-hand side (LHS) into a perfect square form. For (x^2 - 3x):

   • Take half the coefficient of (x): (\frac{-3}{2})
   • Square it: (\left(\frac{-3}{2}\right)^2 = \frac{9}{4})
   • Rewrite the quadratic as:

     x^2 - 3x = \left(x - \frac{3}{2}\right)^2 - \frac{9}{4}
     

3. Isolate the perfect square
   Substitute this back into your equation and move the constant term to the right-hand side (RHS):

   \left(x - \frac{3}{2}\right)^2 = -\frac{t^2}{2} + 2t + C + \frac{9}{4}
   

4. Simplify the RHS and redefine the constant
   To match the square root argument in the given solution, let's eliminate denominators and rearrange terms:

   • Multiply every term by 4 to get rid of fractions:

     4\left(x - \frac{3}{2}\right)^2 = -2t^2 + 8t + 4C + 9
     

   • Factor out (-2) from the (t)-terms: (-2t^2 + 8t = -2t(t-4))
   • Since (C) is an arbitrary constant, we can replace (4C) with a new constant (still called (C) for simplicity—constants can be renamed freely without changing the solution's validity):

     4\left(x - \frac{3}{2}\right)^2 = C - 2t(t-4) + 9
     

5. Solve for (x)
   Take the square root of both sides (we'll use the positive root here; the negative root is another valid solution, which can be accounted for by adjusting the constant (C)):

   2\left(x - \frac{3}{2}\right) = \sqrt{C - 2t(t-4) + 9}
   

   Distribute the 2 on the LHS and rearrange to solve for (x):

   2x - 3 = \sqrt{C - 2t(t-4) + 9}
   x = \frac{3}{2} + \frac{\sqrt{C - 2t(t-4) + 9}}{2}
   

And that's exactly the solution form you were given! The core steps were completing the square for (x) and using the flexibility of arbitrary constants to tidy up the expression inside the square root.

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