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

[MarkFL]

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.

 

[MarkFL]

Hello llllllllllll,

We are given the second order inhomogeneous ODE:

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

Using a power reduction formula, we may write:

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

Observing that the differential operator:

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

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:

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

Computing the derivatives of this solution, we find:

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

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

Substituting into the ODE, we obtain:

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

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

Equating coefficients, we obtain the system:

\(\displaystyle 9A=\frac{1}{2}\implies A=\frac{1}{18}\)

\(\displaystyle B+12C=-\frac{1}{2}\)

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

Thus, the particular solution is:

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