Answer:
So, solution of the differential equation is
![y(t)=-\frac{5x^2}{4}\cot 2x\cdot \cos 2x+c_1e^{-2it}+c_2e^{2it}\\](https://tex.z-dn.net/?f=y%28t%29%3D-%5Cfrac%7B5x%5E2%7D%7B4%7D%5Ccot%202x%5Ccdot%20%5Ccos%20%202x%2Bc_1e%5E%7B-2it%7D%2Bc_2e%5E%7B2it%7D%5C%5C)
Step-by-step explanation:
We have the given differential equation: y′′+4y=5xcos(2x)
We use the Method of Undetermined Coefficients.
We first solve the homogeneous differential equation y′′+4y=0.
![y''+4y=0\\\\r^2+4=0\\\\r=\pm2i\\\\](https://tex.z-dn.net/?f=y%27%27%2B4y%3D0%5C%5C%5C%5Cr%5E2%2B4%3D0%5C%5C%5C%5Cr%3D%5Cpm2i%5C%5C%5C%5C)
It is a homogeneous solution:
![y_h(t)=c_1e^{-2i t}+c_2e^{2i t}](https://tex.z-dn.net/?f=y_h%28t%29%3Dc_1e%5E%7B-2i%20t%7D%2Bc_2e%5E%7B2i%20t%7D)
Now, we finding a particular solution.
![y_p(t)=A5x\cos 2x\\\\y'_p(t)=A5\cos 2x-A10x\sin 2x\\\\y''_p(t)=-A20\sin 2x-A20x\cos 2x\\\\\\\implies y''+4y=5x\cos 2x\\\\-A20\sin 2x-A20x\cos 2x+4\cdot A5x\cos 2x=5x\cos 2x\\\\-A20\sin 2x=5x\cos 2x\\\\A=-\frac{x}{4} \cot 2x\\](https://tex.z-dn.net/?f=y_p%28t%29%3DA5x%5Ccos%202x%5C%5C%5C%5Cy%27_p%28t%29%3DA5%5Ccos%202x-A10x%5Csin%202x%5C%5C%5C%5Cy%27%27_p%28t%29%3D-A20%5Csin%202x-A20x%5Ccos%202x%5C%5C%5C%5C%5C%5C%5Cimplies%20y%27%27%2B4y%3D5x%5Ccos%202x%5C%5C%5C%5C-A20%5Csin%202x-A20x%5Ccos%202x%2B4%5Ccdot%20A5x%5Ccos%202x%3D5x%5Ccos%202x%5C%5C%5C%5C-A20%5Csin%202x%3D5x%5Ccos%202x%5C%5C%5C%5CA%3D-%5Cfrac%7Bx%7D%7B4%7D%20%5Ccot%202x%5C%5C)
we get
![y_p(t)=A5\cos 2x\\\\y_p(t)=-\frac{5x^2}{4}\cot 2x\cdot \cos 2x\\\\\\y(t)=y_p(t)+y_h(t)\\\\y(t)=-\frac{5x^2}{4}\cot 2x\cdot \cos 2x+c_1e^{-2it}+c_2e^{2it}\\](https://tex.z-dn.net/?f=y_p%28t%29%3DA5%5Ccos%202x%5C%5C%5C%5Cy_p%28t%29%3D-%5Cfrac%7B5x%5E2%7D%7B4%7D%5Ccot%202x%5Ccdot%20%5Ccos%20%202x%5C%5C%5C%5C%5C%5Cy%28t%29%3Dy_p%28t%29%2By_h%28t%29%5C%5C%5C%5Cy%28t%29%3D-%5Cfrac%7B5x%5E2%7D%7B4%7D%5Ccot%202x%5Ccdot%20%5Ccos%20%202x%2Bc_1e%5E%7B-2it%7D%2Bc_2e%5E%7B2it%7D%5C%5C)
So, solution of the differential equation is
![y(t)=-\frac{5x^2}{4}\cot 2x\cdot \cos 2x+c_1e^{-2it}+c_2e^{2it}\\](https://tex.z-dn.net/?f=y%28t%29%3D-%5Cfrac%7B5x%5E2%7D%7B4%7D%5Ccot%202x%5Ccdot%20%5Ccos%20%202x%2Bc_1e%5E%7B-2it%7D%2Bc_2e%5E%7B2it%7D%5C%5C)
Answer:
√(p²-4q)
Step-by-step explanation:
Using the Quadratic Formula, we can say that
x = ( -p ± √(p²-4(1)(q))) / 2(1) with the 1 representing the coefficient of x². Simplifying, we get
x = ( -p ± √(p²-4q)) / 2
The roots of the function are therefore at
x = ( -p + √(p²-4q)) / 2 and x = ( -p - √(p²-4q)) / 2. The difference of the roots is thus
( -p + √(p²-4q)) / 2 - ( ( -p - √(p²-4q)) / 2)
= 0 + 2 √(p²-4q)/2
= √(p²-4q)
The average rate, in miles per hour, of the rail portion of the trip is; D: 75
<h3>How to calculate the average speed?</h3>
Since exactly 1/3 of the distance travelled was by rail, then we can say that;
(1/3) * (90) = 30 miles
We are told that part of the trip took 1/5 of the travel time. Thus;
Time it takes to travel that distance is:
(1/5) * (2) = 2/5 of an hour
Average rate is;
30/(2/5) = 150/2 = 75 mph
Read more about Average speed at; brainly.com/question/4931057
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Answer:
Step-by-step explanation:
I don't think u can tell from that because u need the length of each movie
Answer:
C 3 cm per present
Step-by-step explanation:
1 = 3cm
2 = 6cm
3 = 9cm