Understanding the Differential Equation Problem

We are asked to find the solution of the differential equation given by:$$\frac{{dy}}{{dx}} = \cos \left( {y - x} \right) + 1$$

This is a first-order differential equation. Notice the term \(y - x\) inside the cosine function. This structure often suggests using a substitution to simplify the equation.

Solving the Differential Equation using Substitution

Let's introduce a new variable, say \(v\), such that:

$$v = y - x$$

Now, we need to find the derivative of \(v\) with respect to \(x\). Differentiating both sides of the substitution equation with respect to \(x\), we get:

$$\frac{{dv}}{{dx}} = \frac{{d}}{{dx}}(y - x)$$

Using the linearity of differentiation:

$$\frac{{dv}}{{dx}} = \frac{{dy}}{{dx}} - \frac{{dx}}{{dx}}$$

$$\frac{{dv}}{{dx}} = \frac{{dy}}{{dx}} - 1$$

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

$$\frac{{dy}}{{dx}} = \frac{{dv}}{{dx}} + 1$$

Now, substitute \(v = y - x\) and \(\frac{{dy}}{{dx}} = \frac{{dv}}{{dx}} + 1\) into the original differential equation:

$$\frac{{dv}}{{dx}} + 1 = \cos(v) + 1$$

Subtracting 1 from both sides simplifies the equation significantly:

$$\frac{{dv}}{{dx}} = \cos(v)$$

This is now a separable differential equation, where we can separate the variables \(v\) and \(x\).

Separating Variables and Integration

To separate the variables, we move all terms involving \(v\) to one side and all terms involving \(x\) (and \(dx\)) to the other side:

$$\frac{{dv}}{{\cos(v)}} = dx$$

Since \(\frac{1}{\cos(v)} = \sec(v)\), the equation becomes:

$$\sec(v) dv = dx$$

Now, we integrate both sides of the equation:

$$\int \sec(v) dv = \int dx$$

We know the standard integral of \(\sec(v)\). One common form is \(\int \sec(v) dv = \ln |\sec(v) + \tan(v)| + C_1\). However, to match the structure of the given options, let's use another form of the integral or manipulate the result.

Recall that \(\sec(v) = \frac{1}{\cos(v)}\) and \(\tan(v) = \frac{\sin(v)}{\cos(v)}\). So, \(\sec(v) + \tan(v) = \frac{1 + \sin(v)}{\cos(v)}\).

Alternatively, we can use the integral form \(\int \sec(v) dv = -\ln |\sec(v) - \tan(v)| + C_1\). Let's use this form:

$$-\ln |\sec(v) - \tan(v)| = x + C_1$$

where \(C_1\) is the constant of integration.

Multiply by -1:

$$\ln |\sec(v) - \tan(v)| = -x - C_1$$

Exponentiate both sides using base \(e\):

$$|\sec(v) - \tan(v)| = e^{-x - C_1}$$

$$|\sec(v) - \tan(v)| = e^{-x} e^{-C_1}$$

Let \(C\) be a new constant that absorbs \(e^{-C_1}\) and the \(\pm\) sign from the absolute value, i.e., \(C = \pm e^{-C_1}\). Then:

$$\sec(v) - \tan(v) = C e^{-x}$$

Substituting Back and Final Solution

Now, substitute back \(v = y - x\) into the equation:

$$\sec(y - x) - \tan(y - x) = C e^{-x}$$

To match the format of the options, multiply both sides by \(e^x\):

$$e^x [\sec(y - x) - \tan(y - x)] = C e^{-x} e^x$$

$$e^x [\sec(y - x) - \tan(y - x)] = C e^0$$

$$e^x [\sec(y - x) - \tan(y - x)] = C$$

This is the general solution to the given differential equation.

Checking Options

Let's compare our derived solution with the given options:

Option 1: \(e^x[\sec (y – x) – \tan (y – x)] = c\)

Option 2: \(e^x[\sec (y – x) + \tan (y – x)] = c\)

Option 3: \(e^x\sec (y – x) \tan (y – x) = c\)

Option 4: \(e^x = c \sec (y – x) \tan (y – x)\)

Our solution matches Option 1.

Revision Table: Key Concepts in Differential Equations

Additional Information: Integrating Secant Function

The integral of the secant function, \(\int \sec(u) du\), is a common integral in calculus. There are multiple ways to derive or express its result.

One standard formula is:

$$\int \sec(u) du = \ln |\sec(u) + \tan(u)| + C$$

Another valid form, which was useful in this problem, is related:

$$\int \sec(u) du = -\ln |\sec(u) - \tan(u)| + C$$

These forms are related because \(\sec(u) - \tan(u) = \frac{1}{\sec(u) + \tan(u)}\). So, \(-\ln |\sec(u) - \tan(u)| = -\ln \left| \frac{1}{\sec(u) + \tan(u)} \right| = -(\ln(1) - \ln |\sec(u) + \tan(u)|) = 0 - (-\ln |\sec(u) + \tan(u)|) = \ln |\sec(u) + \tan(u)|\). The difference is absorbed into the constant of integration \(C\).

Knowing different forms of standard integrals can sometimes make it easier to match a derived solution to given options in multiple-choice questions.