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

  • Thread starter Thread starter MarkFL
  • Start date Start date
  • Tags Tags
    Ode
AI Thread Summary
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.
MarkFL
Gold Member
MHB
Messages
13,284
Reaction score
12
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.
 
Mathematics news on Phys.org
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:

$$y_p(x)=\frac{1}{18}-\frac{1}{290}\cos(2x)-\frac{6}{145}\sin(2x)$$
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Thread 'Imaginary Pythagoras'
I posted this in the Lame Math thread, but it's got me thinking. Is there any validity to this? Or is it really just a mathematical trick? Naively, I see that i2 + plus 12 does equal zero2. But does this have a meaning? I know one can treat the imaginary number line as just another axis like the reals, but does that mean this does represent a triangle in the complex plane with a hypotenuse of length zero? Ibix offered a rendering of the diagram using what I assume is matrix* notation...
Back
Top