General Approach for Equations with Two First-Derivative Terms #

Hey there! When you're dealing with a differential equation that has y' popping up in multiple spots, here's a straightforward workflow to find its general solution:

• Step 1: Consolidate all y' terms
  Shift every term containing y' to one side of the equation, factor out y', and rearrange things into the form y' * A(x,y) = B(x,y). This lets you isolate y' as a function of x and y, turning the problem into a more familiar type of ordinary differential equation.
• Step 2: Classify the simplified equation
  Once you have y' = f(x,y), figure out which category it falls into:
  • Separable: Split variables and integrate both sides directly.
  • First-order linear: Use the integrating factor method to solve it.
  • Homogeneous: Substitute u = y/x to convert it into a separable equation.
• Step 3: Integrate and add the constant of integration
  After solving the simplified equation, don't forget to include the arbitrary constant C—this gives you the general solution. You can rearrange it into explicit form if possible, but implicit solutions are totally acceptable too.

Solving the Specific Equation: y' = 2(xy' + y)y³ #

Let's work through this equation step by step using the method above:

1. Consolidate y' terms
   First expand the right-hand side:

   y' = 2xy'y³ + 2y⁴
   

   Move all y' terms to the left and factor out y':

   y' - 2xy'y³ = 2y⁴
   y'(1 - 2xy³) = 2y⁴
   

   Isolate y' to get:

   y' = 2y⁴ / (1 - 2xy³)
   

2. Switch variables (treat x as a function of y)
   The expression for y' looks a bit messy, so let's flip the perspective: let x = x(y) (meaning we treat x as a function of y instead of the other way around). This makes y' = 1/x' (where x' denotes dx/dy). Substitute this into the equation:

   1/x' = 2y⁴ / (1 - 2xy³)
   

   Take the reciprocal of both sides and rearrange to get a linear ODE in terms of x:

   x' = (1 - 2xy³)/(2y⁴)
   x' + (1/y)x = 1/(2y⁴)
   

3. Solve the linear ODE with integrating factor
   The integrating factor for a linear equation x' + P(y)x = Q(y) is μ(y) = e^(∫P(y)dy). Here, P(y) = 1/y, so:

   μ(y) = e^(∫(1/y)dy) = e^(ln|y|) = |y|
   

   We can use μ(y) = y (assuming y > 0; the final solution will cover all sign cases anyway). Multiply both sides of the linear equation by y:

   yx' + x = 1/(2y³)
   

   Notice the left-hand side is just the derivative of xy with respect to y:

   d(xy)/dy = 1/(2y³)
   

4. Integrate both sides
   Integrate each side with respect to y:

   ∫d(xy) = ∫1/(2y³) dy
   xy = (1/2)(-1/(2y²)) + C
   xy = -1/(4y²) + C
   

5. Final general solution
   Multiply through by 4y² to eliminate denominators (and absorb the constant into a new arbitrary constant C for simplicity):

   4xy³ + 1 = Cy²
   

   This is the implicit general solution to the original differential equation.

内容的提问来源于stack exchange，提问作者user490308