The solution to the given differential equation is yp=−14xcos(2x)
The characteristic equation for this differential equation is:
P(s)=s2+4
The roots of the characteristic equation are:
s=±2i
Therefore, the homogeneous solution is:
yh=c1sin(2x)+c2cos(2x)
Notice that the forcing function has the same angular frequency as the homogeneous solution. In this case, we have resonance. The particular solution will have the form:
yp=Axsin(2x)+Bxcos(2x)
If you take the second derivative of the equation above for yp , and then substitute that result, y′′p , along with equation for yp above, into the left-hand side of the original differential equation, and then simultaneously solve for the values of A and B that make the left-hand side of the differential equation equal to the forcing function on the right-hand side, sin(2x) , you will find:
A=0
B=−14
Therefore,
yp=−14xcos(2x)
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Answer:
i would say the 3rd answer
