Question

Solve the differential equation \( (x+y)\,dy + (x-y)\,dx = 0 \), given that \( y(1) = 1 \).

Hint:

The equation is homogeneous: put y = vx, separate the variables, integrate, and then use y(1) = 1.

Answer 1

Step 1: Write the equation as dy/dx:
From \( (x+y)\,dy + (x-y)\,dx = 0 \) we get the form below.
\[ \frac{dy}{dx} = \frac{y-x}{x+y} \]
Both numerator and denominator have degree 1, so the equation is homogeneous.

Step 2: Substitute y = vx:
Put \( y = vx \), so \( \frac{dy}{dx} = v + x\frac{dv}{dx} \).
\[ v + x\frac{dv}{dx} = \frac{v-1}{v+1} \]

Step 3: Separate the variables:
Move \( v \) to the right side and combine.
\[ x\frac{dv}{dx} = \frac{v-1}{v+1} - v = \frac{v-1-v^2-v}{v+1} = -\frac{v^2+1}{v+1} \]
\[ \frac{v+1}{v^2+1}\, dv = -\frac{dx}{x} \]

Step 4: Integrate both sides:
Split the left side as \( \frac{v}{v^2+1} + \frac{1}{v^2+1} \).
\[ \frac{1}{2}\log(v^2+1) + \tan^{-1} v = -\log |x| + C \]
Bring \( \log x \) to the left and put \( v = \frac{y}{x} \).
\[ \frac{1}{2}\log\left(\frac{x^2+y^2}{x^2}\right) + \log x + \tan^{-1}\frac{y}{x} = C \]
Since \( \frac{1}{2}\log\frac{x^2+y^2}{x^2} + \log x = \frac{1}{2}\log(x^2+y^2) \), the general solution is:
\[ \frac{1}{2}\log(x^2+y^2) + \tan^{-1}\frac{y}{x} = C \]

Step 5: Use y(1) = 1 to find C:
Put \( x = 1, y = 1 \).
\[ C = \frac{1}{2}\log 2 + \tan^{-1}(1) = \frac{1}{2}\log 2 + \frac{\pi}{4} \]

Final Answer:
The particular solution is given below.
\[ \boxed{\frac{1}{2}\log(x^2+y^2) + \tan^{-1}\frac{y}{x} = \frac{1}{2}\log 2 + \frac{\pi}{4}} \]