To find the general solution of the given differential equation \(\frac{{dy}}{{dx}} = \cos \left( {x + y} \right)\), we will employ a substitution method. This technique helps transform complex differential equations into simpler, often separable, forms.

Differential Equation Substitution

The given differential equation is:

\[ \frac{{dy}}{{dx}} = \cos \left( {x + y} \right) \]

This equation has the term \((x + y)\) inside the cosine function. A common strategy for such forms is to introduce a new variable for this combined term. Let's define a new variable \(v\):

• Let \(v = x + y\).

Next, we need to find the derivative of \(v\) with respect to \(x\), which is \(\frac{{dv}}{{dx}}\). Differentiating both sides of \(v = x + y\) with respect to \(x\):

\[ \frac{{dv}}{{dx}} = \frac{d}{{dx}}\left( {x + y} \right) \]

\[ \frac{{dv}}{{dx}} = 1 + \frac{{dy}}{{dx}} \]

From this expression, we can isolate \(\frac{{dy}}{{dx}}\) to substitute it back into the original differential equation:

\[ \frac{{dy}}{{dx}} = \frac{{dv}}{{dx}} - 1 \]

Substituting into the Differential Equation

Now, we substitute the expressions for \((x + y)\) and \(\frac{{dy}}{{dx}}\) back into the original differential equation \(\frac{{dy}}{{dx}} = \cos \left( {x + y} \right)\):

\[ \frac{{dv}}{{dx}} - 1 = \cos v \]

To proceed, we need to separate the variables. First, move the constant term to the right side of the equation:

\[ \frac{{dv}}{{dx}} = 1 + \cos v \]

Separation of Variables for the Differential Equation

With the equation in the form \(\frac{{dv}}{{dx}} = f(v)\), we can now separate the variables \(v\) and \(x\). This means gathering all terms involving \(v\) on one side and all terms involving \(x\) on the other side:

\[ \frac{{dv}}{{1 + \cos v}} = dx \]

Integration for the General Solution

To find the general solution, we must integrate both sides of the separated equation:

\[ \int {\frac{{dv}}{{1 + \cos v}}} = \int {dx} \]

For the left-hand side integral, we use a fundamental trigonometric identity: \(1 + \cos \theta = 2{\cos ^2}\left( {\frac{\theta }{2}} \right)\). Applying this identity with \(\theta = v\):

\[ \int {\frac{{dv}}{{2{{\cos }^2}\left( {\frac{v}{2}} \right)}}} = \int {dx} \]

We know that \(\frac{1}{{{\cos }^2}\alpha} = {\sec ^2}\alpha\). So, the integral becomes:

\[ \int {\frac{1}{2}{{\sec }^2}\left( {\frac{v}{2}} \right)dv} = \int {dx} \]

Now, we perform the integration for both sides:

• The integral of \(\frac{1}{2}{{\sec }^2}\left( {\frac{v}{2}} \right)\) with respect to \(v\) is \(\frac{1}{2} \cdot \frac{{\tan \left( {\frac{v}{2}} \right)}}{{1/2}} = \tan \left( {\frac{v}{2}} \right)\).
• The integral of \(1\) with respect to \(x\) is \(x\).

After integrating both sides, we introduce the constant of integration, \(c\):

\[ \tan \left( {\frac{v}{2}} \right) = x + c \]

Final General Solution

The final step is to substitute back the original expression for \(v\), which is \(v = x + y\), into our integrated equation. This gives us the general solution of the differential equation in terms of \(x\) and \(y\):

\[ \tan \left( {\frac{{x + y}}{2}} \right) = x + c \]

This equation represents the general solution of the given differential equation.