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 don't know if you wanted me to solve for C or F. But, I Solved for both.
Step-by-step explanation:
<u>Solved for F </u>
<u>Solved for C</u>
Answer:
W = O/U
Step-by-step explanation:
U/1 = O/W
WU=O
W = O/U
Answer:
x = 10 , -12
Step-by-step explanation:
Solution:-
- The given quadratic equation is to be solved using the quadratic formula. The general form of a quadratic equation is:
Where, [ a , b and c are constants ]
- The quadratic formula is given as:
- The given equation is:
Where, a = 1 , b = 2 , c = -120
- Solve using quadratic formula:
