I feel like it would be 1/12 of a chance
Our approach to answering this question is to eliminate the choices until we are left with only one.
(1) FALSE. The given figures are lines and can extend indefinitely.
(2) FALSE. The lines are not given to intersect in right angles.
(3) TRUE. The angles are vertical (which means that they line in opposite side of the intersection.
(4) FALSE. The angles are not supplementary because the lines are vertical. They can only be supplementary if both are right angles.
Answer:
a.![y(x)=c_1e^{2x}+c-2xe^{2x}+x^3e^{2x}](https://tex.z-dn.net/?f=y%28x%29%3Dc_1e%5E%7B2x%7D%2Bc-2xe%5E%7B2x%7D%2Bx%5E3e%5E%7B2x%7D)
b![y(x)=c_1cos 3x+c_2 sin 3x-5 cos x+ 7sin x](https://tex.z-dn.net/?f=y%28x%29%3Dc_1cos%203x%2Bc_2%20sin%203x-5%20cos%20x%2B%207sin%20x)
Step-by-step explanation:
1.![y''-4y'+4y=6x e^{2x}](https://tex.z-dn.net/?f=y%27%27-4y%27%2B4y%3D6x%20e%5E%7B2x%7D)
Auxillary equation
![D^2-4D+ 4=0](https://tex.z-dn.net/?f=D%5E2-4D%2B%204%3D0)
![(D-2)(D-2)=0](https://tex.z-dn.net/?f=%28D-2%29%28D-2%29%3D0)
D=2,2
Then complementary solution =![C_1e^{2x}+C_2xe^{2x}](https://tex.z-dn.net/?f=%20C_1e%5E%7B2x%7D%2BC_2xe%5E%7B2x%7D)
Particular solution ![=\frac{6 xe^{2x}}{(D-2)^2}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B6%20xe%5E%7B2x%7D%7D%7B%28D-2%29%5E2%7D)
D is replace by D+2 then we get
P.I=![\frac{6xe^{2x}}{0}](https://tex.z-dn.net/?f=%5Cfrac%7B6xe%5E%7B2x%7D%7D%7B0%7D)
P.I=![\frac{e^{ax}}{D+a} \cdot .V](https://tex.z-dn.net/?f=%5Cfrac%7Be%5E%7Bax%7D%7D%7BD%2Ba%7D%20%5Ccdot%20.V)
where V is a function of x
P.I=![\frac}x^3e^{2x}](https://tex.z-dn.net/?f=%5Cfrac%7Dx%5E3e%5E%7B2x%7D)
By integrating two times
Hence, the general solution
![y(x)=c_1e^{2x}+c-2xe^{2x}+x^3e^{2x}](https://tex.z-dn.net/?f=y%28x%29%3Dc_1e%5E%7B2x%7D%2Bc-2xe%5E%7B2x%7D%2Bx%5E3e%5E%7B2x%7D)
b.y''+9y=5 cos x-7 sin x
Auxillary equation
![D^2+9=0](https://tex.z-dn.net/?f=D%5E2%2B9%3D0)
D=![\pm 3i](https://tex.z-dn.net/?f=%5Cpm%203i)
![C.F=c_1 cos 3x+ c_2sin 3x](https://tex.z-dn.net/?f=C.F%3Dc_1%20cos%203x%2B%20c_2sin%203x)
P.I=![\frac{5 cos x-7 sin x}{D^2+9}](https://tex.z-dn.net/?f=%5Cfrac%7B5%20cos%20x-7%20sin%20x%7D%7BD%5E2%2B9%7D)
P.I=![\frac{sin ax}{D^2+bD +C}](https://tex.z-dn.net/?f=%5Cfrac%7Bsin%20ax%7D%7BD%5E2%2BbD%20%2BC%7D)
Then D square is replace by -a square
is replace by - then we get
P.I=-5 cos x+7 sin x
The general solution
![y(x)=c_1cos 3x+c_2 sin 3x-5 cos x+ 7sin x](https://tex.z-dn.net/?f=y%28x%29%3Dc_1cos%203x%2Bc_2%20sin%203x-5%20cos%20x%2B%207sin%20x)
24+3.1+(-44)+(8.2)+63
27.1+(-44)+(8.2)+63
(-16.9)+(8.2)+63
(-8.7)+63
54.3