Answer:
![y(x)=e^{-2x}[3cos(\sqrt{6}x)+\frac{2\sqrt{6}}{3}sin(\sqrt{6}x)]](https://tex.z-dn.net/?f=y%28x%29%3De%5E%7B-2x%7D%5B3cos%28%5Csqrt%7B6%7Dx%29%2B%5Cfrac%7B2%5Csqrt%7B6%7D%7D%7B3%7Dsin%28%5Csqrt%7B6%7Dx%29%5D)
(See attached graph)
Step-by-step explanation:
To solve a second-order homogeneous differential equation, we need to substitute each term with the auxiliary equation 
where the values of 
are the roots:
Since the values of are complex conjugate roots, then the general solution is ![y(x)=e^{\alpha x}[C_1cos(\beta x)+C_2sin(\beta x)]](https://tex.z-dn.net/?f=y%28x%29%3De%5E%7B%5Calpha%20x%7D%5BC_1cos%28%5Cbeta%20x%29%2BC_2sin%28%5Cbeta%20x%29%5D)
where 
.
Thus, the general solution for our given differential equation is ![y(x)=e^{-2x}[C_1cos(\sqrt{6}x)+C_2sin(\sqrt{6}x)]](https://tex.z-dn.net/?f=y%28x%29%3De%5E%7B-2x%7D%5BC_1cos%28%5Csqrt%7B6%7Dx%29%2BC_2sin%28%5Csqrt%7B6%7Dx%29%5D)
.
To account for both initial conditions, take the derivative of 
, thus, ![y'(x)=-2e^{-2x}[C_1cos(\sqrt{6}x+C_2sin(\sqrt{6}x)]+e^{-2x}[-C_1\sqrt{6}sin(\sqrt{6}x)+C_2\sqrt{6}cos(\sqrt{6}x)]](https://tex.z-dn.net/?f=y%27%28x%29%3D-2e%5E%7B-2x%7D%5BC_1cos%28%5Csqrt%7B6%7Dx%2BC_2sin%28%5Csqrt%7B6%7Dx%29%5D%2Be%5E%7B-2x%7D%5B-C_1%5Csqrt%7B6%7Dsin%28%5Csqrt%7B6%7Dx%29%2BC_2%5Csqrt%7B6%7Dcos%28%5Csqrt%7B6%7Dx%29%5D)
Now, we can create our system of equations given our initial conditions:
![y(x)=e^{-2x}[C_1cos(\sqrt{6}x)+C_2sin(\sqrt{6}x)]\\\\y(0)=e^{-2(0)}[C_1cos(\sqrt{6}(0))+C_2sin(\sqrt{6}(0))]=3\\\\C_1=3](https://tex.z-dn.net/?f=y%28x%29%3De%5E%7B-2x%7D%5BC_1cos%28%5Csqrt%7B6%7Dx%29%2BC_2sin%28%5Csqrt%7B6%7Dx%29%5D%5C%5C%5C%5Cy%280%29%3De%5E%7B-2%280%29%7D%5BC_1cos%28%5Csqrt%7B6%7D%280%29%29%2BC_2sin%28%5Csqrt%7B6%7D%280%29%29%5D%3D3%5C%5C%5C%5CC_1%3D3)
![y'(x)=-2e^{-2x}[C_1cos(\sqrt{6}x+C_2sin(\sqrt{6}x)]+e^{-2x}[-C_1\sqrt{6}sin(\sqrt{6}x)+C_2\sqrt{6}cos(\sqrt{6}x)]\\\\y'(0)=-2e^{-2(0)}[C_1cos(\sqrt{6}(0))+C_2sin(\sqrt{6}(0))]+e^{-2(0)}[-C_1\sqrt{6}sin(\sqrt{6}(0))+C_2\sqrt{6}cos(\sqrt{6}(0))]=-2\\\\-2C_1+\sqrt{6}C_2=-2](https://tex.z-dn.net/?f=y%27%28x%29%3D-2e%5E%7B-2x%7D%5BC_1cos%28%5Csqrt%7B6%7Dx%2BC_2sin%28%5Csqrt%7B6%7Dx%29%5D%2Be%5E%7B-2x%7D%5B-C_1%5Csqrt%7B6%7Dsin%28%5Csqrt%7B6%7Dx%29%2BC_2%5Csqrt%7B6%7Dcos%28%5Csqrt%7B6%7Dx%29%5D%5C%5C%5C%5Cy%27%280%29%3D-2e%5E%7B-2%280%29%7D%5BC_1cos%28%5Csqrt%7B6%7D%280%29%29%2BC_2sin%28%5Csqrt%7B6%7D%280%29%29%5D%2Be%5E%7B-2%280%29%7D%5B-C_1%5Csqrt%7B6%7Dsin%28%5Csqrt%7B6%7D%280%29%29%2BC_2%5Csqrt%7B6%7Dcos%28%5Csqrt%7B6%7D%280%29%29%5D%3D-2%5C%5C%5C%5C-2C_1%2B%5Csqrt%7B6%7DC_2%3D-2)
We then solve the system of equations, which becomes easy since we already know that 
:
Thus, our final solution is:
![y(x)=e^{-2x}[C_1cos(\sqrt{6}x)+C_2sin(\sqrt{6}x)]\\\\y(x)=e^{-2x}[3cos(\sqrt{6}x)+\frac{2\sqrt{6}}{3}sin(\sqrt{6}x)]](https://tex.z-dn.net/?f=y%28x%29%3De%5E%7B-2x%7D%5BC_1cos%28%5Csqrt%7B6%7Dx%29%2BC_2sin%28%5Csqrt%7B6%7Dx%29%5D%5C%5C%5C%5Cy%28x%29%3De%5E%7B-2x%7D%5B3cos%28%5Csqrt%7B6%7Dx%29%2B%5Cfrac%7B2%5Csqrt%7B6%7D%7D%7B3%7Dsin%28%5Csqrt%7B6%7Dx%29%5D)
Answer:
Hose A fills at a rate of 10 gal per min
Hose B fills at a rate of 8 gal per min
Hose B fills at a faster rate
Step-by-step explanation:
1. You have that Chris used 18 gallons of gas in 420 miles.
2. Therefore: If Chris drove 420 miles using 18 gallons of gas, how many gallons would in 357 miles?
3. Let's call "x" to the amount of gallons he need to drive 357 miles. So, you have:
420 miles----18 gallons
357 miles-----x
x=(357)(18)/420
x=6426/420
x=15.3 gallons
Answer:
DE = 21
Step-by-step explanation:
Recall: the ratio of the corresponding side lengths of similar triangles are equal
Given that ∆ABC ~ ∆DEF, therefore,
AB/DE = BC/EF = AC/DF
AB = 8.4
DE = x
BC = 10
EF = 25
AC = 16.5
DF = 41.25
Let's find DE using AB/DE = BC/EF. Thus:
8.4/x = 10/25
Cross multiply
x*10 = 25*8.4
10x = 210
Divide both sides by 10
x = 210/10
x = 21
DE = x = 21
