Solve Calc Problem: Find y if dy/dt=ky and k is a Nonzero Constant

  • Thread starter Thread starter radtad
  • Start date Start date
AI Thread Summary
To solve the differential equation dy/dt = ky, where k is a nonzero constant, the general solution involves integrating to find y = Ce^(kt), where C is the constant of integration. The user initially believed the answer was b (2e^(kt)), but realized they overlooked the constant of integration. The correct form includes a constant, which can be represented as C1, leading to multiple valid solutions depending on initial conditions. A typo in the answer choices was also acknowledged, confirming that choice c should be e^(kt) + 3. Understanding the role of the constant of integration is crucial for identifying the correct solution.
radtad
Messages
19
Reaction score
0
We have a set of problems for hw. I am stuck on 1 where I know the answer but can't seem to get it.

If dy/dt=ky and k is a nonzero constant then y could be
a. 2e^kty b. 2e^kt c. e^kt d. kty+5 e. 1/2ky^2 +1/2

I know the answer is b but i can't get that answer
Here is my work
S=integral sign

dy/dt=ky
dy/y=kdt
Sdy/y=kSdt
lny=kt
e^lny=e^kt
y=e^kt

How do u get a 2 in there for choice b
 
Last edited:
Physics news on Phys.org
You forgot about the constant of integration:
\frac{dy}{dt}=ky
\frac{dy}y=kdt
\int{\frac{dy}y} = \int{kdt}
\ln y = kt +C
e^{\ln y} = e^{kt + C}
y = e^{kt}\cdot e^C
eC is also a constant, so it can be written as C1 if you like.The value of C1 will depend on the initial conditions. Unless there's a typo in your answer list, I can see two answers that are of this form:
b. y=2e^{kt}
and c. y = e^{kt}

I hope that helps.
 
thanks forgot the C and yea choice c was a typo it should be e^kt +3
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...

Similar threads

Back
Top