MHB Find a Particular Solution y_p for 2nd Order ODE at Yahoo Answers

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The discussion focuses on finding a particular solution y_p for the second-order inhomogeneous ordinary differential equation (ODE) 2y'' + 6y' + 9y = sin^2(x) using the Method of Undetermined Coefficients. The equation is transformed using a power reduction formula, yielding a right-hand side of 1/2 - 1/2 cos(2x). The differential operator is identified, and the proposed form for the particular solution is y_p(x) = A + Bcos(2x) + Csin(2x). By substituting the derivatives into the ODE and equating coefficients, the values A = 1/18, B = -1/290, and C = -6/145 are determined, leading to the final particular solution. This method effectively illustrates the process of solving a specific type of inhomogeneous ODE.
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Here is the question:

Find a particular solution y_p of the differential equation 2y′′+6y′+9y=sin^2(x)?


Find a particular solution y_p of the differential equation
2y′′+6y′+9y=sin^2(x).

using the Method of Undetermined Coefficients. Primes denote derivatives with respect to x. Thank you.

I have posted a link there to this topic so the OP can see my work.
 
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Hello llllllllllll,

We are given the second order inhomogeneous ODE:

$$2y''+6y'+9y=\sin^2(x)$$

Using a power reduction formula, we may write:

$$2y′′+6y′+9y=\frac{1}{2}-\frac{1}{2}\cos(2x)$$

Observing that the differential operator:

$$A\equiv D\left(D^2+4 \right)$$

annihilates the RHS of the ODE, and that none of the roots of this operator are the characteristic roots of the associated homogenous ODE, we know the particular solution must take the form:

$$y_p(x)=A+B\cos(2x)+C\sin(2x)$$

Computing the derivatives of this solution, we find:

$$y_p'(x)=-2B\sin(2x)+2C\cos(2x)$$

$$y_p''(x)=-4B\cos(2x)-4C\sin(2x)$$

Substituting into the ODE, we obtain:

$$2\left(-4B\cos(2x)-4C\sin(2x) \right)+6\left(-2B\sin(2x)+2C\cos(2x) \right)+9\left(A+B\cos(2x)+C\sin(2x) \right)=\frac{1}{2}-\frac{1}{2}\cos(2x)$$

$$9A+(B+12C)\cos(2x)+(C-12B)\sin(2x)=\frac{1}{2}+\left(-\frac{1}{2} \right)\cos(2x)+0\cdot\sin(2x)$$

Equating coefficients, we obtain the system:

$$9A=\frac{1}{2}\implies A=\frac{1}{18}$$

$$B+12C=-\frac{1}{2}$$

$$C-12B=0\implies C=12B\implies B=-\frac{1}{290},\,C=-\frac{6}{145}$$

Thus, the particular solution is:

$$y_p(x)=\frac{1}{18}-\frac{1}{290}\cos(2x)-\frac{6}{145}\sin(2x)$$
 
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