**Title: Solving Differential Equations Using the Method of Undetermined Coefficients****Introduction:**In this exercise, we will find a particular solution to the differential equation using the Method of Undetermined Coefficients.**Problem Statement:**Given the differential equation:\[ x''(t) - 14x'(t) + 49x(t) = 3te^{7t} \]**Methodology:**The Method of Undetermined Coefficients is a technique used to find particular solutions to linear differential equations with constant coefficients and specific types of non-homogeneous terms.**Solution Outline:**1. **Identify the form of the particular solution**: Based on the right-hand side of the equation \(3te^{7t}\), propose a suitable form for \(x_p(t)\) that might include terms with similar exponential growth and polynomial functions.2. **Determine undetermined coefficients**: Substitute the proposed solution into the differential equation to determine the unknown coefficients.3. **Combine with homogeneous solution**: The total solution will include both the particular and homogeneous solutions.**Solution:**A solution is given by:\[ x_p(t) = \, \_\_\_\_ \] (Fill in the particular solution form and coefficients once calculated)**Summary:**Once the particular solution \(x_p(t)\) is found, it can be added to any complementary solution of the homogeneous equation to form the general solution to the original problem.**Conclusion:**The Method of Undetermined Coefficients offers a structured way to tackle non-homogeneous linear differential equations, especially when the non-homogeneous part matches specific forms.

The given differential equation is: x''t-14x't+49xt=3te7t. We have to find a particular solution to the given differential equation using the method of undetermined coefficients.We have, x''t-14x't+49xt=3te7t . . . (1) Let us suppose that xpt=t2At+Be7t=At3+Bt2e7t . . . (2) Differentiating (2) w.r.t. t on both sides we get, x'=7At3+Bt2e7t+3At2+2Bte7tx'=7At3+7B+3At2+2Bte7t . . . 3 Again differentiating (3) w.r.t. t on both sides we get, x''=77At3+7B+3At2+2Bte7t+21At2+27B+3At+2Be7tx''=A49t3+42t2+6te7t+B49t2+28t+2e7t . . . 4Now substituting the values of x , x' , and x'' from equations (2), (3), and (4) in equation (1) we get, A49t3+42t2+6te7t+B49t2+28t+2e7t-147At3+7B+3At2+2Bte7t+49At3+Bt2e7t=3te7t 6Ate7t+2Be7t=3te7t Equating the coefficient of like terms we get, A=36=12B=0 Hence the particular solution will be:xpt=t3e7t2