# Anti-Diff and Differential Equations help.

1. May 13, 2007

### Jozsa

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Hey, theres 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

2. May 13, 2007

### Hootenanny

Staff Emeritus
Welcome to the forums,

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

3. May 13, 2007

### cristo

Staff Emeritus
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! ]

4. May 13, 2007

### Hootenanny

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

[best of luck with your exams cristo!]

5. May 13, 2007

### cristo

Staff Emeritus
Sorry... yea sure you can!
Thanks; good luck in your remaining exams!

6. May 13, 2007

### Jozsa

Okay Update,

1.

Find the area enclosed by the curves:

y² = 4 + x and x + 2y = 4

I worked it out myself, so dont 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 im 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?

Last edited: May 13, 2007
7. May 13, 2007

### Jozsa

Anyone? Help!!!