To find the general solution of the given first-order differential equation, we will employ a strategic substitution method. The equation presented is:

\[ \frac{{dy}}{{dx}} = {\left( {x + y - 1} \right)^2} \]

Differential Equation Substitution

The structure of this differential equation, particularly the repeated term \((x + y - 1)\), suggests that a substitution can significantly simplify it. Let's define a new variable, \(z\), to represent this complex expression:

Let \(z = x + y - 1\)

Next, we need to find the derivative of \(z\) with respect to \(x\), which is \(\frac{{dz}}{{dx}}\). Differentiating both sides of our substitution \(z = x + y - 1\) with respect to \(x\):

\[ \frac{{dz}}{{dx}} = \frac{{d}}{{dx}}(x + y - 1) \]

Applying the rules of differentiation (derivative of \(x\) is 1, derivative of \(y\) is \(\frac{{dy}}{{dx}}\), and derivative of a constant is 0):

\[ \frac{{dz}}{{dx}} = 1 + \frac{{dy}}{{dx}} - 0 \]

From this, we can express \(\frac{{dy}}{{dx}}\) in terms of \(\frac{{dz}}{{dx}}\):

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

Transforming the Differential Equation

Now, we substitute \(z\) and our new expression for \(\frac{{dy}}{{dx}}\) back into the original differential equation:

Original equation: \(\frac{{dy}}{{dx}} = {\left( {x + y - 1} \right)^2}\)

Substituting the expressions derived from our substitution:

\[ \frac{{dz}}{{dx}} - 1 = z^2 \]

To solve this new equation, we rearrange it to isolate \(\frac{{dz}}{{dx}}\):

\[ \frac{{dz}}{{dx}} = z^2 + 1 \]

This transformed equation is now a separable differential equation. This means we can separate the variables \(z\) and \(x\) to integrate them independently.

Separating Variables and Integration

To solve the separable differential equation, we move all terms involving \(z\) to one side with \(dz\), and all terms involving \(x\) to the other side with \(dx\):

\[ \frac{{dz}}{{z^2 + 1}} = dx \]

Now, we integrate both sides of the equation:

\[ \int \frac{{dz}}{{z^2 + 1}} = \int dx \]

The integral on the left-hand side is a standard integral form: \(\int \frac{1}{a^2 + u^2} du = \frac{1}{a} \tan^{-1}\left(\frac{u}{a}\right) + C\). In our case, \(a=1\) and \(u=z\). The integral on the right-hand side is straightforward.

Performing the integration on both sides gives:

\[ \tan^{-1}(z) = x + C \]

Here, \(C\) represents the arbitrary constant of integration.

Back Substitution to Find General Solution

The final step to obtaining the general solution in terms of \(x\) and \(y\) is to substitute back the original expression for \(z\), which was \(z = x + y - 1\), into our integrated equation:

\[ \tan^{-1}(x + y - 1) = x + C \]

To explicitly solve for \(y\), we take the tangent of both sides of the equation:

\[ x + y - 1 = \tan(x + C) \]

Finally, rearrange the equation to isolate \(y\):

\[ y = 1 - x + \tan(x + C) \]

This is the general solution for the given first-order differential equation, where \(C\) is a constant.