Let yp(x)=x*sin*3*x is a perticular solution of y'+py'=6*cos*3*x. then p equals?

Question

Answer

Finding the value of p in the given differential equation:

Given:
yp(x) = x*sin(3x) is a particular solution of y' + py' = 6*cos(3x)

Explanation:
To find the value of p, we first differentiate yp(x) twice to get the first and second derivatives.

First derivative:
yp'(x) = sin(3x) + 3x*cos(3x)

Second derivative:
yp''(x) = 6*cos(3x) - 9x*sin(3x)
Substitute yp(x), yp'(x), and yp''(x) into the given differential equation:
6*cos(3x) - 9x*sin(3x) + p(sin(3x) + 3x*cos(3x)) = 6*cos(3x)
Now, simplify the equation by grouping like terms:
(6 - 9p)*cos(3x) + (3p)*x*sin(3x) = 0
For this equation to hold for all x, the coefficients of cos(3x) and x*sin(3x) must be zero. Therefore, we get two equations:
6 - 9p = 0
3p = 0
Solving these equations simultaneously, we find:
p = 6/9
p = 0.
Therefore, the value of p in the given differential equation is p = 2/3.

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