1) yy'-e^x= 0, and y=4 when x=0 means that:
A)y=x-lnx^2+4
B)y^2=4x^2+3
C)y=-2x+1/2x^3
D)y^2=2e^x+14
E)1/2ln|x^2+4|+6
2) The rate of change of P with respect to t is directly proportional to (10-t) and inversely proportional to the cube root of P. The differential equation that models this situation is:
A)dP/dt=k(10-t)/1/3sqrt(p)
B)dP/dt=(10-t)/sqrt(p)
C)dP/dt=k(10-t)3sqrt(p)
D)dP/dt=k3sqrt(p)/10-t
E)dP/dt=k(10-t)/3sqrt(p)
Us Online students gotta stick together. Much love
Answers D and E were correct.
y dy = e^x dx
1/2 y^2 = e^x + c
if (0,4) is on the graph, then
8 = 1+c
c = 7
1/2 y^2 = e^x + 7
Looks like D to me
dP/dt = k(10-t)/∛t
Looks like E to me
If you must use 3 to indicate the root, why say sqrt? The sq clearly implies square root. Try 3root of cbrt.
Jiskha Gang On Foe Nem
Jiskha's been saving my life since quarantine started lol
1) B)y^2=4x^2+3 - Well, you know what they say - when life throws you a differential equation, just square it and add some random numbers!
2) E)dP/dt=k(10-t)/3sqrt(p) - Ah, the differential equation of love! The rate of change of P is directly proportional to (10-t) and inversely proportional to the cube root of P. Who knew math could be so romantic? Well, except for the k value, that's just there to keep things interesting.
To solve these questions, we need to apply basic differential equation concepts and techniques. Let's go through each question step by step.
1) The given differential equation is yy' - e^x = 0, and we are given the initial condition y = 4 when x = 0.
To solve this differential equation, we can separate the variables and integrate both sides.
Start by isolating yy':
yy' = e^x
Next, divide both sides by y:
y'/y = e^x/y
Now, integrate both sides with respect to x:
∫(y'/y) dx = ∫(e^x/y) dx
The left side of the equation simplifies to ln|y|, and the right side simplifies to ∫e^x dx.
So we have:
ln|y| = e^x + C
To find the constant C, we can use the initial condition y = 4 when x = 0:
ln|4| = e^0 + C
ln|4| = 1 + C
Simplifying this equation, we have:
ln(4) = 1 + C
C = ln(4) - 1
Substituting the value of C back into the equation, we get:
ln|y| = e^x + ln(4) - 1
Finally, we can simplify ln|y| using logarithm properties:
ln|y| = ln(4e^x) - 1
ln|y| = ln(4e^x) - ln(e)
ln|y| = ln(4e^x/e)
ln|y| = ln(e^(2x)/e)
ln|y| = ln(e^(2x-1))
|y| = e^(2x-1)
Taking the positive value for y, we have:
y = e^(2x-1)
So the correct answer is A) y = x - ln(x^2) + 4.
2) The rate of change of P with respect to t is directly proportional to (10 - t) and inversely proportional to the cube root of P. We need to find the differential equation that models this situation.
Let's assume the constant of proportionality is k.
Based on the given information, we can write the differential equation as follows:
dP/dt = k(10 - t)/(P^(1/3))
To simplify this equation, we can rearrange as:
(P^(1/3)) * dP = k(10 - t) * dt
Now, integrate both sides of the equation:
∫ (P^(1/3)) * dP = ∫ k(10 - t) * dt
Integrating the respective sides:
(3/4) * P^(4/3) = k * (10t - (t^2)/2) + C
Here, C is the constant of integration.
So, the correct answer is:
C) dP/dt = k * (10 - t) * (P^(1/3))