he rate of change of y with respect to t is 3 times the value of the quantity 2
less than y. Find an equation for y, given that y = 212 when t = 0.
You get:
A. y = 212e^3t - 2
B. y = 210e^3t +2
C. y = 212e^3t +2
D. y= 214e^3t - 2
E. y = 210e^3t - 2
Thank you it was correct :)
ever think of typing real math?
dy/dt = 3(y-2)
dy/(y-2) = 3dt
ln(y-2) = 3t+c
y-2 = c*e^(3t)
Since y(0) = 212, c = 210
y = 210e^(3t)+2
Can you help me with this one, please
The rate of change of y with respect to x is one-half times the value of y. Find an equation for y, given that y = -7 when x = 0.
You get:
A. y=-7e^0.5x
B. y=e^0.5x -7
C. y=-7(1/2)^x
D. y=-7(1/2)e^x
E. dy/dx=1/2*y
dy/dx = y/2
dy/y = (1/2) dx
ln y = x/2 + C
y = c e^(x/2)
y = -7 e^(x/2) which is A
It really so say he
Well, let's break this down. We know that the rate of change of y with respect to t is 3 times the value of the quantity 2 less than y.
If we let "a" represent the quantity 2 less than y, then we can write that the rate of change of y with respect to t is 3a.
Now, if we integrate both sides with respect to t, we get:
∫(dy/dt) dt = ∫3a dt
Integrating the left side gives us y, and integrating the right side gives us 3at + C, where C is a constant.
So now we have:
y = 3at + C
But we also know that y = 212 when t = 0. Plugging in these values gives us:
212 = 3a(0) + C
C = 212
Substituting this value of C back in, we get:
y = 3at + 212
Now, if we multiply through by e^3t, we get:
y = 3a(e^3t) + 212(e^3t)
Since a represents the quantity 2 less than y, we can substitute that in:
y = 3(y - 2)(e^3t) + 212(e^3t)
Expanding and simplifying gives us:
y = 3ye^3t - 6e^3t + 212e^3t
Finally, combining like terms, we get:
y = 3ye^3t + 206e^3t
So, the answer is option:
C. y = 212e^3t + 2
To find the equation for y, we need to start with the given information that the rate of change of y with respect to t is 3 times the value of the quantity 2 less than y. Let's break down the problem step by step:
1. We are given that the rate of change of y with respect to t is 3 times the value of 2 less than y. Mathematically, this can be represented as dy/dt = 3(y - 2).
2. Next, we need to solve this first-order differential equation. We can rearrange the equation as dy/(y - 2) = 3dt.
3. Now, we can integrate both sides of the equation. The integral of dy/(y - 2) can be found using the substitution method. Let's let u = y - 2, then du = dy.
4. The integral becomes ∫du/u = 3∫dt. Solving these integrals gives us ln|u| = 3t + C, where C is the constant of integration.
5. Substituting back u = y - 2, we have ln|y - 2| = 3t + C.
6. We know that y = 212 when t = 0, so we can substitute these values into the equation. ln|212 - 2| = 0 + C. Simplifying gives ln|210| = C.
7. We can rewrite the equation as ln|y - 2| = 3t + ln|210|.
8. By using properties of logarithms, we can combine the logarithms on the right side of the equation: ln|y - 2| = ln|210e^3t|.
9. Applying the property of logarithms that ln(a) = ln(b) implies a = b, we find y - 2 = 210e^3t.
10. Finally, we can solve for y by adding 2 to both sides of the equation, giving us y = 210e^3t + 2.
Therefore, the equation for y, given that y = 212 when t = 0, is y = 210e^3t + 2. Hence, the correct option is B.