Anti-Diff and Differential Equations help.

[Jozsa]

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Hey, there's a few questions i need help with urgently, would apreciate any help on as many of them as fast as possible. Thanks in advance.

1.
Find the area enclosed by the curves:
y² = 4 + x and x + 2y = 4


2.
Fine the volume of the solid of revolution generated by rotating about the y-axis the region enclosed by the y-axis and the curves y = 2x^2 - 1 and y =
√(x)


3.
Use the method of cylindrical shells to find the volume of the solid of revolution generated by rotating about the y-axis the region enclosed by y =
√(x+1), y = 0, x=0 and x=4


Differential equations now:

4.
Solve
dy/dx = (x(y²+3)/y

5.
Solve
dy/dx = e^(x-2y), y(1) = 0

6.
dy/dx = x²y-2e^(x), y(1) = -1

Use Euler method with 4 stephs to find and approx value for y(2)




Thank you sooo much for the help

 

[Hootenanny]

Welcome to the forums,

According to our guidelines, one is expected to show one's own efforts before asking for help.

 

[cristo]

Welcome to PF Jozsa. Please note that you are rquired to show some work before we can help you. So, what have you attempted for the questions thus far?

[Damn.. beaten to it again! I'll just go back to my own revision then! ]

 

[Hootenanny]

[Damn.. beaten to it again! I'll just go back to my own revision then! ]

Your making me feel guilty now... I can revise and post at the same time... right?

[best of luck with your exams cristo!]

 

[cristo]

Your making me feel guilty now... I can revise and post at the same time... right?

Sorry... yea sure you can!

[best of luck with your exams cristo!]

Thanks; good luck in your remaining exams!

 

[Jozsa]

Okay Update,



1.

Find the area enclosed by the curves:

y² = 4 + x and x + 2y = 4



I worked it out myself, so don't worry









2.

Fine the volume of the solid of revolution generated by rotating about the y-axis the region enclosed by the y-axis and the curves y = 2x^2 - 1 and y =

√(x)



V = integral 1 to 0 of ( pi (√(x) - 2x^2 - 1)^2 )



So i expanded (√(x) - 2x^2 - 1)^2



to get (simplified): 4x^4 - 4x^2.5 - 4x^2 + 2√(x) + x + 1



then got the integral is:



pi((4/5)x^5 - (4/3.5)x^3.5 - (4/3)x^3 + (2/1.5)x^1.5 + (x^2)/2 + x) between 0.42 and 0



so subbing in those values: pi((81/70) - 0)



= 3.64 units^3



Correct?







3.

Use the method of cylindrical shells to find the volume of the solid of revolution generated by rotating about the y-axis the region enclosed by y = √(x+1), y = 0, x=0 and x=4



V = integral_{4 to 0} of [ 2pi x ( √(x)+1 ) ] dx



= (2pi)integral_{4 to 0} of (x^1.5 + x) dx



= (2pi)[(x^2.5)/2.5 + (x^2)/2] between 4 and 0



= (2pi)(12.8 + 8)



= 41.6 pi units^3



Correct?





4.

Solve

dy/dx = (x(y²+3)/y)



(y/(y^2+3)) dy/dx = x



so: integral (y/(y^2+3)) dy = integral x dx + C



so the integral is: (1/2)ln(y^2+3) = x^2/2 + C



then, we say ln(y^2+3) = x^2 + A

where A = 2C



so, y^2 = = e^(x^2 + A) - 3



y = √(e^(x^2 + A) - 3)



Correct?







5.

Solve

dy/dx = e^(x-2y), y(1) = 0



ok.. so I am pretty much stumped after this step



ln(dy) - ln(dx) = x - 2y



ln(dy) + 2y = ln(dx) + x



Help?







6.

dy/dx = x²y-2e^(x), y(1) = -1

Use Euler method with 4 steps to find and approx value for y(2)



The formula for calculating succcessive values of x and y is :

y_n+1 = y_n + 0.25[(x_n^2)(y_n)-2e^(x_n)]



(_n+1 means subscript n+1 etc)



y_1 = -1 + 0.25[(1^2)(-1)-2e^(1)] = -2.6091

y_2 = -2.6091 + 0.25[(1.25^2)(-2.6091)-2e^(1.25)] = -5.3735

y_3 = -5.3735 + 0.25[(1.50^2)(-5.3735)-2e^(1.50)] = -10.6369

y_4 = -10.6369 + 0.25[(1.75^2)(-10.6369)-2e^(1.75)] = -21.6581





So the value at y(2) = -21.6581



Yes? Correct?

 

[Jozsa]

Anyone? Help!