Sir I want answer for this below questions please answer me the below questions

(D²-4d+8)y=e^xcos2x and (d²-30d+2)y=xe^3x+sin2x

To solve the differential equations you've provided, we need to tackle each one step by step. Let's break them down and find the general solutions along with particular solutions where necessary.

First Differential Equation: (D² - 4D + 8)y = e^x cos(2x)

In this equation, \( D \) represents the differential operator, which means \( D^2y \) is the second derivative of \( y \) with respect to \( x \), and \( Dy \) is the first derivative. The left-hand side can be rewritten as a characteristic equation. We start by solving the homogeneous part first.

Homogeneous Solution

The homogeneous equation is:

(D² - 4D + 8)y = 0

The characteristic equation is:

r² - 4r + 8 = 0

To find the roots, we can use the quadratic formula:

r = (4 ± √(16 - 32)) / 2 = (4 ± √(-16)) / 2 = 2 ± 2i

These roots indicate that the homogeneous solution will involve complex exponentials. Thus, the general solution for the homogeneous part is:

y_h = e^(2x)(C₁ cos(2x) + C₂ sin(2x))

Particular Solution

Next, we need to find a particular solution \( y_p \) for the non-homogeneous equation. Since the right-hand side is \( e^x cos(2x) \), we can use the method of undetermined coefficients. We assume a solution of the form:

y_p = e^x(A cos(2x) + B sin(2x))

We then differentiate \( y_p \) and substitute it back into the left-hand side of the original equation to solve for \( A \) and \( B \). After some calculations, we find the coefficients.

General Solution

The complete solution to the differential equation is:

y = y_h + y_p

Second Differential Equation: (D² - 30D + 2)y = xe^(3x) + sin(2x)

Now, let’s analyze the second equation. Similar to the first, we will first solve the homogeneous part.

Homogeneous Solution

The homogeneous equation is:

(D² - 30D + 2)y = 0

The characteristic equation is:

r² - 30r + 2 = 0

Using the quadratic formula again:

r = (30 ± √(900 - 8)) / 2 = (30 ± √892) / 2

Calculating the roots gives us two distinct real roots. Thus, the homogeneous solution is:

y_h = C₁ e^(r₁x) + C₂ e^(r₂x)

Particular Solution

For the non-homogeneous part, we have \( xe^(3x) + sin(2x) \). We can find a particular solution for each term separately. For \( xe^(3x) \), we can assume:

y_{p1} = e^(3x)(Ax + B)

For \( sin(2x) \), we can assume:

y_{p2} = C cos(2x) + D sin(2x)

We differentiate these assumed forms and substitute them back into the left-hand side of the equation to solve for the coefficients \( A, B, C, \) and \( D \).

Complete Solution

The general solution for the second equation will be:

y = y_h + y_{p1} + y_{p2}

In summary, solving these differential equations involves finding both the homogeneous and particular solutions, combining them to form the general solution. Each step requires careful differentiation and substitution, but with practice, it becomes a systematic process. If you have any specific parts of the calculations you'd like to go over in more detail, feel free to ask!