Recent content by Illusionist

Homework Statement Consider the second order differential equation y'' - 4y' + 4y = f(x) Find a particular solution if f(x) = 25cos(x) Homework Equations I believe for this type of question I should let y = Asin(x) + B cos(x) Hence y' = Acos(x) - Bsin(x) and y'' = -Asin(x) -...

Very true. Well I gave that a try and this is how I went: lim [(-2/n^2)sin(2/n)] / (-1/n^2) lim 2sin(2/n) 2sin (lim [2/n]) 2sin(0) 0, hence convergent. Please tell me I'm right! Thank you so much for the replies guys.

Of course but I'm a bit stumped as to what to do with this one. Any suggestions? I presume that means what I did before applying L'Hopital's was right. Thanks for the reply.

Homework Statement Determine whether the following sequence, whose nth term is given, converges or diverges. Find the limit of each convergent one. n[1 - cos(2/n)] Homework Equations I have made a solid attempt and obtained an answer but I am convinced I made a mistake and have missed...

Homework Statement Find the general solution of the following differential equation: x.(dy/dx) = y + sqrt.[(x^2) - (y^2)] Homework Equations I'm working through my excerise book and have been able to get through quite a few differential equations with success, but this one really does...

Yeah I get what you mean matness, thanks for everything mate. I just not really confident with what I do and why, like in this question. Still not sure what I did was right for the horizontal strips.

Yeah sorry. Between 0 to 1 would make sense, as both horizontal and vertical would give me an area of (Pi/2) now, but using the same sort of approach is my horizontal method still right? Thanks for that matness, and yeah I really should stop relying so heavily on my calculator.

Homework Statement Find the area of the region bounded by the positive x-axis, the positive y-axis and the curve: (x^2) +[(y^2)/4] = 1 using vertical and horizontal strips. Homework Equations Basically I just tried to use integration to find the area, but I suspect I have made a...

OK yeah C = 8A - 13 I recalculated: 10= 4Ax - (3 - A)x + x(8A-13) 10= 3Ax 16x A = (16/13) + (10/13)x This value of x gave me the following for B and C: B = 3 - [(16/13) + (10/13)x] = (23/13) - (10/13)x C = 8[(16/13) + (10/13)x] - 13 C = (80/13)x - (41/13) Now I still think something is...

Alright here is what I have come up with: (3x^2 + 10x + 13) = A(x^2 + 4x + 8) + (Bx + C)(x-1) (3x^2 + 10x + 13) = A(x^2) + 4xA + 8A + B(x^2) - Bx + Cx - C Hence - (x^2): 3 = A + B -------> B = 3 - A (x^1): 10 = 4A - B + C (x^0): 13 = 8A - C...

Homework Statement [(3x^2)+10x+13]/[(x-1)([x^2]+4x+8)] Homework Equations I think solving this question should include partial fractions. The Attempt at a Solution I've made a few different attempts at this question but find myself at a dead end every time. One attempt was...

Please, anyone?

Would anyone know where I went wrong above?

Ok so I'm at V''(x^4) + 4(x^3)V'=10(x^7) + 15(x^2). I then just devided all by (x^4) to get V'' alone, hence" V'' + V'(4/x) = 10(x^3) + 15/(x^2) I then let u=V' and u'=V'', therefore: u' +u(4/x) = 10(x^3) + 15/(x^2), I then let P(x)=4/x and hence I(x)=x^4 after integration. Now I have (d/dx)...

Homework Statement Given that y=x^2 is a solution to the differential equation: (x^2)y'' + 2xy' - 6y = 0 <--- Eq.(1) find the general solution of the differential equation (x^2)y'' + 2xy' - 6y = 10(x^7) + 15(x^2) <--- Eq.(2) Hence write...