Solve the following differential equations :
`(x+y)dy+(x-y)dx=0`,
given that `y=1` when `x=1`

Answer

To solve the differential equation \((x+y)dy + (x-y)dx = 0\) with the initial condition \(y=1\) when \(x=1\), we will follow these steps:

### Step 1: Rewrite the Equation
We start with the equation:
\[ (x+y)dy + (x-y)dx = 0 \]
We can rearrange this to isolate \(dy\):
\[ (x+y)dy = -(x-y)dx \]
Dividing both sides by \((x+y)(x-y)\) gives:
\[ \frac{dy}{dx} = -\frac{x-y}{x+y} \]

### Step 2: Check for Homogeneity
To check if the equation is homogeneous, we can substitute \(x = \lambda x\) and \(y = \lambda y\):
\[ \frac{dy}{dx} = -\frac{\lambda x - \lambda y}{\lambda x + \lambda y} = -\frac{x - y}{x + y} \]
Since the equation remains unchanged, it is homogeneous.

### Step 3: Substitute \(y = vx\)
Let \(y = vx\), where \(v\) is a function of \(x\). Then, we have:
\[ dy = vdx + x\frac{dv}{dx} \]
Substituting \(y\) and \(dy\) into the equation gives:
\[ (x + vx)(vdx + x\frac{dv}{dx}) + (x - vx)dx = 0 \]
This simplifies to:
\[ (x + vx)(vdx + x\frac{dv}{dx}) + (x - vx)dx = 0 \]

### Step 4: Simplify the Equation
Expanding and simplifying:
\[ (x + vx)vdx + (x + vx)x\frac{dv}{dx} + (x - vx)dx = 0 \]
Combining like terms leads to:
\[ x^2\frac{dv}{dx} + v(x + vx) = -\frac{(x - vx)}{(x + vx)} \]

### Step 5: Separate Variables
Rearranging gives us:
\[ \frac{dv}{dx} = -\frac{(x - vx)}{x^2(1 + v)} \]
This can be separated as:
\[ \frac{(1 + v)}{(v - 1)} dv = -\frac{1}{x^2} dx \]

### Step 6: Integrate Both Sides
Integrating both sides:
\[ \int \frac{(1 + v)}{(v - 1)} dv = -\int \frac{1}{x^2} dx \]
The left side can be integrated using partial fractions, while the right side gives:
\[ -\frac{1}{x} + C \]

### Step 7: Solve for \(v\)
After integrating, we will solve for \(v\) and substitute back \(v = \frac{y}{x}\).

### Step 8: Apply Initial Condition
The initial condition \(y = 1\) when \(x = 1\) will help us find the constant \(C\).

### Final Solution
After applying the initial condition, we will arrive at the final solution in the form:
\[ \tan^{-1}\left(\frac{y}{x}\right) + \frac{1}{2}\log(x^2 + y^2) = C \]