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Ashok Kumar
Ashok Kumar, Electrical Engineer
Category: Math Homework
Satisfied Customers: 7307
Experience:  Master's Degree
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Can you help me with these questions

This answer was rated:

Hello can you help me with these questions

Are only answers needed?

Customer: replied 6 months ago.

OK. I will reply shortly.

q1(a) y = ln(x^2-3)

(b) solution of y' = 2y/(x^2-1)

y = 6(x-1)/(x+1)

(c) solution of (3-x) dy/dx = y

y = 5/(3-x)

(d) solution of dy/dx = 2x(1+y^2)

y = tan)(x^2-1)

Customer: replied 6 months ago.
Thank you and can you also help me with questions 2 the second page

yes, I am working and will upload answers shortly.

q2(a)

differential eqn

dh/dt = k*sqrt(h)

2sqrt(h) = kt+c

at t=0, h =0 gives c = 0.

hence, 2sqrt(h)=kt

at t=2 , h =9, hence, 2sqrt(9) = 2k or, k =3

hence, 2sqrt(h) = 3t, or, h =(3/2)^2t^2= 2.25t^2

h = 2.25t^2

h= 36, t=?

or, 36 = 2.25t^2 or, f =sqrt(36/2.25)=4

Hence, after 4 hrs, height of water will be 4 m in the tank.

q2(b) we have

dP/dt = -kP

OR, dP/P = -kt

or, lnP = -kt+c or, P = Ae^(-kt). where e^c= A

At t = 0, P = 10000; hence, 10000= A

Hence, P = 10000*e^-kt

At t = 10, P = 4*10000/5, hence, 4*10000/5 = 10000*e^(-10k)

or, k = -0.1ln(0.8) = 0.022

2000=10000e^(-0.022t)

or, t = 73.156, say 74

After 74 days, the population will reduce below 2000.

Customer: replied 6 months ago.
Thank you so much

you are most welcome.

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