Question

A particular integral of the differential equation 
\[ y'' + 3y' + 2y = e^{e^x} \] is 
 

• A. \( e^{e^x} e^{-x} \)
• B. \( e^{e^x} e^{-2x} \)
• C. \( e^{e^x} e^{2x} \)
• D. \( e^{e^x} e^x \)

Hint:

In solving non-homogeneous differential equations, use undetermined coefficients or variation of parameters to find the particular solution.

Answer 1

The Correct Option is B

To find a particular integral for the given differential equation:

\(y'' + 3y' + 2y = e^{e^x}\)

we can use the method of undetermined coefficients or variation of parameters. Given the nature of the function on the right-hand side, \(e^{e^x}\), we'll use an intelligent guess to find the particular integral.

The homogeneous equation is:

\(y'' + 3y' + 2y = 0\)

The characteristic equation for this is:

\(m^2 + 3m + 2 = 0\)

Factoring or solving this quadratic equation, we get:

\((m + 1)(m + 2) = 0\)

The roots are \(m = -1\) and \(m = -2\).

Therefore, the complementary function (CF) is:

\(y_c = C_1 e^{-x} + C_2 e^{-2x}\)

For the particular integral (PI), notice that \(e^{e^x}\) is not a common function associated with standard forms. However, we see this as a candidate for solution: \(y_p = A e^{e^x} e^{\beta x}\). Let's determine \(\beta\).

Assuming a form, let's choose \(y_p = A e^{e^x} e^{bx}\), and substitute into the left side of the differential equation.

Substituting \(y_p = A e^{e^x} e^{-2x}\) into the differential equation will allow the left side to equal \(e^{e^x}\) correctly.

Thus, the particular integral is:

\(y_p = e^{e^x} e^{-2x}\)

Hence, the correct answer is:

Option 2: \( e^{e^x} e^{-2x} \)