Anti-Differentiation and Differential Equations

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?

Quote:

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Originally Posted by Jozsa

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

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First sketch the curves, see attachment.

Now to find this area it is easier to integrate in the y-direction, so write
the curves in the form x=f(y), to get:

a) x = y^2-4
b) x = 4 - 2y

Then the area is:

A = integral_{y=-4 to 2} (4 - 2y)-(y^2-4) dy

which is elementary and you should be able to do it yourself
(its 36, which is what I get by counting squares:cool: ).

RonL

Quote:

---

Originally Posted by Jozsa

4.
Solve
dy/dx = (x(y²+3)/y

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This is of variables seperable type, so we have:

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

so:

integral [y/(y^2+3)] dy = integral x dx + C

or:

(1/2)ln(y^2+3) = x^2/2 + C

and I will leave you to simplify this.

RonL

Thanks RonL for the help so far!

I actually managed to work out the first one myself, turns out i just made a simple mistake the first time i tried to do it, but it was good to see that your answer backed up mine.


as for question 4, im having some trouble simplifying.

(1/2)ln(y^2+3) = x^2/2 + C


I think i may have it

Firstly, we say ln(y^2+3) = x^2 + A
Where A = 2C

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

Finally, y = √(e^(x^2 + A) - 3)

Is this correct?

Would appreciate some help on the other questions too!

For Question 2, i got 0.35 units^3
Feedback?

Quote:

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Originally Posted by CaptainBlack

integral [y/(y^2+3)] dy = integral x dx + C

or:

(1/2)ln(y^2+3) = x^2/2 + C

---

I may be wrong... but isnt (1/2)ln(y^2+3) the integral of 1/(y^2+3)?
The question stated that it is y/(y^2+3) not 1/(y^2+3)

I could be, and probably am wrong :) but just making sure.

Quote:

---

Originally Posted by Jozsa

I may be wrong... but isnt (1/2)ln(y^2+3) the integral of 1/(y^2+3)?
The question stated that it is y/(y^2+3) not 1/(y^2+3)

I could be, and probably am wrong :) but just making sure.

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d/dy[1/(y^2+3)] = -2y/(y^2+3)^2

could you help me out with 5 and 6, im totally stuck.

and let me know if ive gone about the other questions correctly.

thanks!

I think i worked out Q6, still stuck on 5 though, REALLY need help.

Quote:

---

Originally Posted by Jozsa

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

---

dy/dx = e^x e^{-2y)

so:

e^{2y}dy/dx = e^x

hence:

integral e^{2y} dy = integal e^x dx + C

(1/2) e^{2y} = e^x + C

RonL

Quote:

---

Originally Posted by CaptainBlack

dy/dx = e^x e^{-2y)

so:

e^{2y}dy/dx = e^x

hence:

integral e^{2y} dy = integal e^x dx + C

(1/2) e^{2y} = e^x + C

RonL

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Youre my hero, thank you so much friend.
Sorry for suddenly joining and being a pest :(
But i really honestly do appreciate it.

Also, could you verify that my solutions to the other problems are correct?

Thank you.