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:5/18=.2777777777777
Step-by-step explanation:
.27777 (7 repeating)
100x=27.7777
10x= 2.7777
Subtract
90x=25
x=25/90=5/18
Check
5/18=.2777777777777
Hope that helps.
Answer:
31.1
Step-by-step explanation:
Answer:
Class Boundary = 1 between the sixth and seventh classes.
Step-by-step explanation:
Lengths (mm) Frequency
1. 140 - 143 1
2. 144 - 147 16
3. 148 - 151 71
4. 152 - 155 108
5. 156 - 159 83
6. 160 - 163 18
7. 164 - 167 3
The class boundary between two classes is the numerical value between the starting value of the higher class, which is 164 for the 7th class in this case, and the ending value of the class of the lower class, which is 163 for the 6th class in this case.
Therefore the class boundary between the sixth and seventh classes
= 164 - 163 = 1
Therefore Class Boundary = 1.
It can be seen that class boundary for the frequency distribution is 1.
If we take the difference between the lower limits of one class and the lower limit of the next class then we will get the class width value.
Therefore, Class width,
w = lower limit of second class - lower limit of first class
= 144 - 140
= 4
