Solve the following differential equations :
`cos(x+y)dy=dx, y(0)=0`.

Answer

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

### Step 1: Rewrite the equation
We start with the given equation:
\[ \cos(x+y) \, dy = dx \]
We can rearrange this to isolate \( dy \) on one side:
\[ dy = \frac{dx}{\cos(x+y)} \]

### Step 2: Introduce a substitution
Let \( v = x + y \). Then, we have:
\[ y = v - x \quad \text{and thus} \quad dy = dv - dx \]
Substituting \( dy \) into the equation gives:
\[ dv - dx = \frac{dx}{\cos(v)} \]
Rearranging this, we get:
\[ dv = dx \left(1 + \frac{1}{\cos(v)}\right) \]

### Step 3: Separate variables
Now, we can separate the variables:
\[ \frac{dv}{1 + \frac{1}{\cos(v)}} = dx \]
This simplifies to:
\[ \frac{\cos(v) \, dv}{\cos(v) + 1} = dx \]

### Step 4: Integrate both sides
Now we will integrate both sides. The left side requires a substitution:
\[ \int \frac{\cos(v) \, dv}{\cos(v) + 1} = \int dx \]
The right side integrates to \( x + C \).

### Step 5: Solve the left integral
To solve the left integral, we can use the identity:
\[ \cos(v) + 1 = 2\cos^2\left(\frac{v}{2}\right) \]
Thus, the integral becomes:
\[ \int \frac{\cos(v) \, dv}{2\cos^2\left(\frac{v}{2}\right)} = \int \frac{1}{2} \sec^2\left(\frac{v}{2}\right) \, dv \]
Integrating gives:
\[ \frac{1}{2} \tan\left(\frac{v}{2}\right) + C \]

### Step 6: Combine results
Setting the results from both sides equal gives:
\[ \frac{1}{2} \tan\left(\frac{v}{2}\right) = x + C \]
Substituting back \( v = x + y \):
\[ \frac{1}{2} \tan\left(\frac{x+y}{2}\right) = x + C \]

### Step 7: Apply initial condition
Using the initial condition \( y(0) = 0 \):
\[ \frac{1}{2} \tan\left(\frac{0 + 0}{2}\right) = 0 + C \implies C = 0 \]
Thus, we have:
\[ \frac{1}{2} \tan\left(\frac{x+y}{2}\right) = x \]

### Step 8: Solve for y
Rearranging gives:
\[ \tan\left(\frac{x+y}{2}\right) = 2x \]
To find \( y \):
\[ \frac{x+y}{2} = \tan^{-1}(2x) \implies x + y = 2\tan^{-1}(2x) \implies y = 2\tan^{-1}(2x) - x \]

### Final Solution
The solution to the differential equation is:
\[ y = 2\tan^{-1}(2x) - x \]