Question

Find a particular solution of the differential equation \( (x - y) (dx + dy) = dx - dy \) when \( y = -1 \) if \( x = 0 \).

Hint:

When a differential equation contains terms like \( (x-y) \) or \( (x+y) \), it's a strong hint to try a substitution like \( v = x-y \) or \( v = x+y \). This often transforms the equation into a much simpler, separable form.

Answer 1

Step 1: Understanding the Concept:
The given differential equation is not immediately in a standard form. We need to rearrange it to identify its type. After rearranging, it becomes a homogeneous differential equation that can be solved using the substitution \( v = x - y \). Once the general solution is found, we use the given initial conditions to find the particular solution.
Step 2: Key Formula or Approach:
1. Rearrange the equation to the form \( \frac{dy}{dx} = f(x, y) \).
2. Use the substitution \( v = x - y \), which implies \( \frac{dv}{dx} = 1 - \frac{dy}{dx} \) or \( \frac{dy}{dx} = 1 - \frac{dv}{dx} \).
3. Solve the resulting separable equation in \( v \) and \( x \).
4. Substitute back \( v = x - y \) to get the general solution.
5. Apply the initial condition \( y(0) = -1 \) to find the integration constant.
Step 3: Detailed Explanation or Calculation:
1. Rearrange the Equation:
\[ (x - y) (dx + dy) = dx - dy \] \[ x dx + x dy - y dx - y dy = dx - dy \] Group terms with \( dy \) and \( dx \):
\[ x dy - y dy + dy = dx - x dx + y dx \] \[ (x - y + 1) dy = (1 - x + y) dx \] \[ \frac{dy}{dx} = \frac{1 - (x - y)}{1 + (x - y)} \] 2. Substitute \( v = x - y \):
Let \( v = x - y \). Then \( \frac{dy}{dx} = 1 - \frac{dv}{dx} \). The equation becomes:
\[ 1 - \frac{dv}{dx} = \frac{1 - v}{1 + v} \] \[ \frac{dv}{dx} = 1 - \frac{1 - v}{1 + v} = \frac{(1 + v) - (1 - v)}{1 + v} = \frac{1 + v - 1 + v}{1 + v} = \frac{2v}{1 + v} \] 3. Solve the Separable Equation:
\[ \frac{1 + v}{2v} dv = dx \] Integrate both sides:
\[ \int \frac{1 + v}{2v} dv = \int dx \] \[ \int \left( \frac{1}{2v} + \frac{1}{2} \right) dv = \int dx \] \[ \frac{1}{2}\ln|v| + \frac{1}{2}v = x + C_0 \] Multiply by 2:
\[ \ln|v| + v = 2x + 2C_0 \] Let \( C = 2C_0 \).
\[ \ln|v| + v = 2x + C \] 4. Substitute Back:
Replace \( v \) with \( x - y \):
\[ \ln|x - y| + (x - y) = 2x + C \] \[ \ln|x - y| = x + y + C \] This is the general solution.
5. Apply Initial Condition:
We are given \( y = -1 \) when \( x = 0 \). Substitute these values to find C:
\[ \ln|0 - (-1)| = 0 + (-1) + C \] \[ \ln|1| = -1 + C \] \[ 0 = -1 + C \] \[ C = 1 \] Step 4: Final Answer:
Substituting \( C=1 \) back into the general solution gives the particular solution:
\[ \ln|x - y| = x + y + 1 \]