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:
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180-2(55)-x=0
180-110-x=0
70-x=0
-x=-70
x=70°
Answer:
Step-by-step explanation:
Given expression is,
(2x - 1)² + 2(2x - 1) = (2x - 1)(2x + 1)
To prove this identity we will take the left hand side of the equation and will prove equal to the right side.
(2x - 1)² + 2(2x - 1) = (2x - 1)(2x + 1)
4x² - 4x + 1 + 4x - 2 = (2x - 1)(2x + 1)
4x² - 1 = (2x - 1)(2x + 1)
(2x - 1)(2x + 1) = (2x - 1)(2x + 1) [Since a² - b² = (a - b)(a + b)]
