Recent content by dan5

yes yp is the particular solution. Well i thought of using a trial particular solution of yp= A sin (Ht) + B cos (Ht) which leaves me with -AH2-24BH+432A = 0 for coefficients of sin(Ht) and -BH2+24AH+432B = 1 for coefficients of cos(Ht)

Homework Statement y'' + 24y' +432y = cos (wt) yp(x) = R(h) cos (ht - ∅(h)) find R(h) and hbar = value for h which R(h) is max. and ∅(hbar) Homework Equations double angle formula for cos cos (a-b) = cos a cos b + sin a sin b The Attempt at a Solution solving the...

Ahhh now I see, thanks to you, and to Euler!

e-z2 where z is a complex number a+ib

Thanks for that idea, I have done that and am left with y=e0.5t ( i 2 sin (√7/2)t ) But as I am meant to be plotting y as a function of t, I don't see how the imaginary part will factor in

How do you go about sketching y as a function of t for t≥0 y= e(0.5t + i(√7/2)t) - e(0.5t-i(√7/2)t) I know it goes through the origin, and the gradient is positive here. But I'm unsure on how to deal with the imaginary numbers when I have a graph of y vs t.

Right, here goes... Integrating WRT y 2∫1 [5x2y+(2/3)y3 ] then entering limits 2∫1 [5x3+(2/3)x3] - [5x2(2-x) + (2/3)(2-x)3] simplifying to... 2∫1 (32/3)x3 - 10x2 - (2/3)(2-x)3 applying the second integral WRT x [ (32/12) x4 - (10/3) x3 + (1/6)(2-x)4 ] then...

Thanks for the help... but when I then integrate WRT y I get 2∫1 [5x2y+(2/3) y3] which I have to substitute in my limits of y=x and y=2-x but then this leaves me with a (2/3) (2-x)3 term, which does not simplify well, and does not lead me to the correct answer. Any ideas where...

Evaluate ∫∫R 5x2 + 2y2 where R is triangle (1,1) (2,0) (2,2) I see the lines bounding the triangle are y=x y=2-x and x=2, and have tried many attempts at setting up the correct limits. Would it be correct to split this into 2 triangles, or are the limits y=x∫y=2-x for y and 2∫1...