Question

Solve d^2 y /dx^2 + 9y = xcos2x

Answer 1

Answer:

Taking the first and second derivatives of y_p with respect to x, we get:

y_p' = A cos(2x) + 2Bx cos(2x) - 2Ax sin(2x) + B sin(2x),

y_p'' = -2A sin(2x) - 4Ax cos(2x) - 4Bx sin

Step-by-step explanation:

To solve this differential equation, we can use the method of undetermined coefficients. First, we find the general solution of the homogeneous equation d^2 y /dx^2 + 9y = 0:

The characteristic equation is r^2 + 9 = 0, which has the roots r = ±3i. Therefore, the general solution of the homogeneous equation is:

y_h = c1 cos(3x) + c2 sin(3x),

where c1 and c2 are constants of integration.

Next, we need to find a particular solution of the non-homogeneous equation d^2 y /dx^2 + 9y = xcos(2x). Since the right-hand side of the equation is a product of a polynomial and a trigonometric function, we can assume a particular solution of the form:

y_p = Ax^2 cos(2x) + Bx sin(2x) + C cos(2x) + D sin(2x),

where A, B, C, and D are constants to be determined.

Taking the first and second derivatives of y_p with respect to x, we get:

y_p' = 2Ax cos(2x) - 2Ax^2 sin(2x) + B cos(2x) + 2Bx cos(2x) + 2C sin(2x) + 2D cos(2x),

y_p'' = -4Ax sin(2x) - 4Ax cos(2x) - 4Ax^2 cos(2x) + 2B sin(2x) + 4Bx sin(2x) - 4C cos(2x) + 4D sin(2x).

Substituting y_p, y_p', and y_p'' into the non-homogeneous equation, we get:

(-4Ax sin(2x) - 4Ax cos(2x) - 4Ax^2 cos(2x) + 2B sin(2x) + 4Bx sin(2x) - 4C cos(2x) + 4D sin(2x)) + 9(Ax^2 cos(2x) + Bx sin(2x) + C cos(2x) + D sin(2x)) = x cos(2x).

Matching the coefficients of like terms on both sides of the equation, we get the following system of equations:

-4A + 9C = 0,

-4A + 4B + 9D = 0,

-4A - 4C = 0,

2B = 1,

4B = 0.

The last equation is inconsistent, so we cannot find a particular solution of the form y_p = Ax^2 cos(2x) + Bx sin(2x) + C cos(2x) + D sin(2x)).

Instead, we need to try a different form for y_p. Since the right-hand side of the equation is a product of a polynomial and a trigonometric function, we can assume a particular solution of the form:

y_p = x(A cos(2x) + B sin(2x)),

where A and B are constants to be determined.

Taking the first and second derivatives of y_p with respect to x, we get:

y_p' = A cos(2x) + 2Bx cos(2x) - 2Ax sin(2x) + B sin(2x),

y_p'' = -2A sin(2x) - 4Ax cos(2x) - 4Bx sin

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