Question

Find the general solution to the differential equation y''(x) -3y'(x) +2y(x) = eˣ sinx

Answer 1

To find the general solution to the given differential equation, first solve the associated homogeneous equation. Then, find a particular solution using the method of undetermined coefficients. Finally, combine the complementary and particular solutions to obtain the general solution.

Step-by-step explanation:

To find the general solution to the given differential equation, we first find the complementary solution by solving the associated homogeneous equation: y''(x) -3y'(x) +2y(x) = 0. The characteristic equation for this equation is r^2 -3r +2 = 0, which factors as (r-1)(r-2) = 0. Therefore, the complementary solution is y_c(x) = C_1e^x + C_2e^2x, where C_1 and C_2 are constants.

To find a particular solution to the non-homogeneous equation, we can use the method of undetermined coefficients. Since the right side of the equation is in the form e^x sin(x), we guess a particular solution of the form y_p(x) = Ae^x sin(x) + Be^x cos(x). By substituting this into the differential equation, we can determine the values of A and B.

Therefore, the general solution to the given differential equation is y(x) = y_c(x) + y_p(x), where y_c(x) is the complementary solution and y_p(x) is the particular solution.

Answer 2

The general solution to the given differential equation is
`\( y''(x) - 3y'(x) + 2y(x) = e^x \sin x \).`

To find the general solution to the differential equation
`\( y''(x) - 3y'(x) + 2y(x) = e^x \sin x \),`we'll follow these steps:

Step 1: Solve the Homogeneous Equation

First, we solve the homogeneous part of the differential equation, which is:

`\[ y''(x) - 3y'(x) + 2y(x) = 0 \]`

This is a second-order linear homogeneous differential equation with constant coefficients. The characteristic equation is:

`\[ r^2 - 3r + 2 = 0 \]`

Step 2: Find the Roots of the Characteristic Equation

We find the roots of the characteristic equation to determine the solution of the homogeneous equation.

Step 3: Solve the Non-homogeneous Equation

After obtaining the solution to the homogeneous equation, we find a particular solution to the non-homogeneous equation
`\( y''(x) - 3y'(x) + 2y(x) = e^x \sin x \)` using an appropriate method like undetermined coefficients or variation of parameters.

Step 4: Combine the Solutions

The general solution to the differential equation will be the sum of the homogeneous solution and the particular solution.

Let's start by solving the homogeneous part.

The solution to the homogeneous differential equation
`\( y''(x) - 3y'(x) + 2y(x) = 0 \) is:`

`\[ y_h(x) = (C_1 + C_2e^x)e^x \]`

where
`\( C_1 \) and \( C_2 \)` are constants.

Step 3: Solve the Non-homogeneous Equation

Now, we need to find a particular solution to the non-homogeneous differential equation
`\( y''(x) - 3y'(x) + 2y(x) = e^x \sin x \). Given the right-hand side is \( e^x \sin x \),`we'll use the method of undetermined coefficients to find a particular solution. We guess a solution of the form:

`\[ y_p(x) = e^x(A \cos x + B \sin x) \]`

where A and B are coefficients to be determined.

Step 4: Determine A and B

We'll differentiate
`\( y_p(x) \) to get \( y_p'(x) \) and \( y_p''(x) \)`, substitute them into the differential equation, and equate the coefficients of like terms to find A and B

Let's proceed with these calculations.

The values of the coefficients A and B in the particular solution
`\( y_p(x) = e^x(A \cos x + B \sin x) \) are \( A = (1)/(2) \) and \( B = -(1)/(2) \).` Therefore, the particular solution is:

`\[ y_p(x) = e^x\left((1)/(2) \cos x - (1)/(2) \sin x\right) \]`

Step 4: Combine the Solutions

The general solution to the differential equation is the sum of the homogeneous solution and the particular solution:

`\[ y(x) = y_h(x) + y_p(x) = (C_1 + C_2e^x)e^x + e^x\left((1)/(2) \cos x - (1)/(2) \sin x\right) \]`