Understanding the Differential Equation and Particular Integral

The given differential equation is of the form \(f(D)y = e^{ax}\), where \(f(D) = D^2+2D-1\). We are also given that a particular integral (PI) for this equation is \(\frac{{ - 4}}{7}{e^{ax}}\).

To find the particular integral for a differential equation of the form \(f(D)y = e^{ax}\), we typically use the formula:

\(PI = \frac{1}{f(D)}e^{ax}\)

For the exponential function \(e^{ax}\) on the right-hand side, if \(f(a) \neq 0\), the particular integral is found by substituting \(D\) with \(a\) in the operator \(f(D)\):

\(PI = \frac{1}{f(a)}e^{ax}\)

In our case, \(f(D) = D^2+2D-1\). So, we substitute \(D=a\):

\(f(a) = a^2+2a-1\)

Assuming \(f(a) \neq 0\), the particular integral is:

\(PI = \frac{1}{a^2+2a-1}e^{ax}\)

Equating the Calculated PI with the Given PI

We are given that the particular integral is \(\frac{{ - 4}}{7}{e^{ax}}\). We can now equate our calculated PI with the given PI:

\(\frac{1}{a^2+2a-1}e^{ax} = \frac{{ - 4}}{7}{e^{ax}}\)

Since \(e^{ax}\) is a common term (and is not zero), we can equate the coefficients:

\(\frac{1}{a^2+2a-1} = \frac{{ - 4}}{7}\)

Solving for the Value of 'a'

Now we need to solve this equation for 'a'. We can cross-multiply:

\(1 \times 7 = -4 \times (a^2+2a-1)\)

\(7 = -4a^2 - 8a + 4\)

Rearrange the terms to form a quadratic equation:

\(4a^2 + 8a + 7 - 4 = 0\)

\(4a^2 + 8a + 3 = 0\)

This is a quadratic equation of the form \(Ax^2 + Bx + C = 0\), where \(A=4\), \(B=8\), and \(C=3\). We can solve for 'a' using the quadratic formula:

\(a = \frac{{ - B \pm \sqrt{{B^2} - 4AC} }}{{2A}}\)

Substitute the values of A, B, and C:

\(a = \frac{{ - 8 \pm \sqrt{{8^2} - 4(4)(3)} }}{{2(4)}}\)

\(a = \frac{{ - 8 \pm \sqrt{{64} - 48} }}{8}\)

\(a = \frac{{ - 8 \pm \sqrt{16} }}{8}\)

\(a = \frac{{ - 8 \pm 4 }}{8}\)

We get two possible values for 'a':

• \(a_1 = \frac{{ - 8 + 4 }}{8} = \frac{{ - 4 }}{8} = -\frac{1}{2}\)
• \(a_2 = \frac{{ - 8 - 4 }}{8} = \frac{{ - 12 }}{8} = -\frac{3}{2}\)

We should also verify that for these values of 'a', \(f(a) = a^2+2a-1 \neq 0\). As calculated before, for \(a = -1/2\) or \(a = -3/2\), \(a^2+2a-1 = -7/4 \neq 0\). So the method used for finding PI is valid for these values of 'a'.

Thus, the possible values for 'a' are \(-\frac{1}{2}\) and \(-\frac{3}{2}\).

Comparing these values with the given options, we find that they match option 2.

Summary of Steps

1. Identify the differential operator \(f(D)\) and the RHS \(e^{ax}\).
2. Recall the method for finding the particular integral \(PI = \frac{1}{f(D)}e^{ax}\).
3. Evaluate \(f(a)\) and form the PI as \(\frac{1}{f(a)}e^{ax}\).
4. Equate the calculated PI with the given PI.
5. Solve the resulting equation for 'a' (which is a quadratic equation).
6. Check if \(f(a) \neq 0\) for the obtained values of 'a'.

The values of 'a' that satisfy the condition are \(-\frac{1}{2}\) and \(-\frac{3}{2}\).