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 one
Step-by-step explanation:
Given three points, it is possible to draw a circle that passes through all three. The perpendicular bisectors of a chords always passes through the center of the circle. By this method we find the center and can then draw the circle. This is virtually the same as constructing the circumcircle a triangle.
Answer:
B
Step-by-step explanation:
I'm not sure if it's right but you can try it
Answer:
missing info my dude
Step-by-step explanation:
