OpenStudy (anonymous):

Write down a second order inhomogeneous differential equation whose homogeneous cousin has a solution that is a linear combination of exp(x)sin(2x) and exp(x)cos(2x), while the original equation has RHS: 3xexp(x)sin(2x)+5xexp(x)cos(2x)+7x^3. What is the shape of the general solution of the inhomogenous differential equation?

3 years ago
OpenStudy (anonymous):

You want to find an equation such that \[y_h=C_1e^x\sin2x+C_2e^x\cos2x\] is the homogeneous solution. You'll get this sort of solution if the roots to the characteristic equation are \(r=1\pm2i\): \[(r-(1+2i))(r-(1-2i))=0~~\iff~~r^2-2r+5=0\] which has the differential-equation form, \[y''-2y'+5y=0\] You're told that the nonhomogeneous part of the equation is \[3xe^x\sin2x+5xe^x\cos2x+7x^3\] so the original DE was \[y''-2y'+5y=3xe^x\sin2x+5xe^x\cos2x+7x^3\]

3 years ago
OpenStudy (anonymous):

oh my gosh you made that look so easy. Is that all there is to it? Maybe I just have a hard time understanding what the question is asking me. Thank you!!!!!

3 years ago
OpenStudy (anonymous):

You're welcome!

3 years ago
OpenStudy (anonymous):

so to figure out the shape do I just need to try and graph it?

3 years ago