Anyone good at symbolic differential equations?
85418
dx/dt+ax=b :P\n\n\nmathematica had some online find-the-derivitive program, last I remember
765615
differentiate implicitly:\n\n2x + 2yy' = 0\n\nand then some algebra\n\ny' = -(x/y)\n\nthen differentiate again:\n\ny'' = (-y - (-x)*y')/y^2 = (-y + xy')/y^2\n\nthen substitute in values:\n\ny' = -0/5 = 0\n\nand y'' = (-5 + 0*0)/(5^2) = =-5/25 = -1/5
299918
This is a separable differential equation\ndP/dt = 4P^2 + tP^2\nFactor P^2 out from the left side\ndP/dt = p^2(4+t)\nDivide each side by P^2 and multiply each side by dt\ndp/P^2 = (4+t)dt\nNow integrate each side\n-1/P = 4t + (1/2)t^2 + C\nSolve for P and you will get\nP = -1/(4t + (1/2)t^2 + C)
794869
Here's a good website that discuss implicit differentiation with some example problems.\n\nHope this helps
929562
∫ax(2 -x)dx\n= ∫(2ax - ax²)dx\n= 2a * ½x² - a * ⅓x³ + c\n= ax² - ⅓ax³ + c
750522
Begin with where you left of:\n\n2x + 2yy' = 0\n\nDifferentiate again, just like the first time.  Use the chain rule on the yy' term:\n\n2 + 2(y'y' + yy'') = 0\n1 + y'^2 + yy'' = 0\ny'' = (-1-y'^2)/ y\n\nFrom your y' = -x/y\n\ny'' = (-1-(-x/y)^2)/ y = -x^2/y^3 - 1/y\n\nThis isn't the answer you have but see if they are the same:\n\nx^2 = 4 - y^2\n\ny'' = -(4-y^2)/y^3 - 1/y = -4/y^3 + 1/y - 1/y = -4/y^3\n\nSo the two solutions are the same
49770
There is no closed form for the nth derivative of this function.  I symbolically calculated and factored the first 50 derivatives on MathCAD, and there was not a pattern.  Sorry, but you're out of luck....
796480
use v^2 =  GM/r\n\nv = root GM/2r\n\norbit = 2 pi r\n\nfor part 2 use suvat equations, s=vu+.5at^2 etc and it should fall out
865054
Let u=ln t, e^u=t, du=dt/t, dt=t du\nThen:\n∫ln² t dt = ∫t u² du = ∫u² e^u du\nIntegrating by parts:\nu² e^u - ∫2u e^u du\nu² e^u - 2u e^u + ∫2e^u du\nu² e^u - 2u e^u + 2e^u + C\nt ln² t - 2t ln t + 2t +C\nAnd we are done.
846527
You'll have to do this in more than one step.  Basically it will sort of have to be a program.  I would start with the restart; command and then  define the function (I would look this up in the help menu since it is dependent on 2 variables).  Then use a command to find the derivative.  Then a command to solve for the pairs (x,y) when the derivative is 0.  That will give you 1.  For 2 and 3 you will also have to do a series of commands.  Use your calc book to refresh your memory on the procedure of finding these, it should be a similar procedure in maple.
962855
should be e^u*du/dx\n\nd[e^(x^2)] = e(x^2)*2*x
892157
wait 1 sec.\nOk this is reasonably difficult, you have to use the chain rule and the quotient rule.\nUsing the chain rule:\nderivative = [ (1/2) * ( (2x + 3) / (7x+4) ) ^ (-1/2) ]  *  [d/dx (2x+3) / (7x+4) ]\n\nSo now you to use the quotient rule to find:\nd/dx (2x+3) / (7x+4) = (14x+8 - 14x + 21) / (7x+4)^2\n= 29 / (7x+4)^2\n\nSo the derivative is [1/2 * ((2x+3)/(7x+4))^(-1/2)]*[29 / (7x+4)^2]\n\n\nBy the way i got the thumbs down before I wrote all this up, so it doesn't mean it's wrong.
196268
(x^2+1)dy/dx+2xy=4x^2 \n---> (x^2+1)dy+(2xy-4x^2)dx=0\n\nIt is a perfect differential because:\n\n∂(x^2+1)/∂y=∂(2xy-4x^2)/∂x\n\n\n--->d(yx^2+y-4x^3/3)=0\n\n--->yx^2+y-4x^3/3=c --->\n y= (c+4x^3/3)/(1+x^2)\n\nNow we check answer:\n\ndy/dx= \n[4x^2*(1+x^2)-2x*(c+4x^3/3)]/(1+x^2)^2\n\n---> (x^2+1)dy/dx+2xy = \n\n[4x^2*(1+x^2)-2x*(c+4x^3/3)]/\n(1+x^2)+2x(c+4x^3/3)/(1+x^2)\n\n=[4x^2*(1+x^2)-2x*(c+4x^3/3)+2x(c+4x^3/3)]/\n(1+x^2)\n\n=4x^2   so it is the correct answer.\n\nSince ordinary homogeneous differential equations can\nhave only one general solution, our answer is the one.
771697
Try Google or Yahoo.  You'll find worked examples there.  I think it needs diagrams to understand; they and the accompanying explanations are too complex to attempt to reproduce here.
212357
As I am sure you soon realized, this is a bit challenging. You cannot solve it easily by algebra, although you can make a series expansion approximation of ln y, or exp(y) and approximate the answer to this kind of equation in some cases. \n\nA more general approach is apply some numerical technique, like the Newton-Raphson method. \n\nLet F(Y)=ln(Y) + 10Y -4\n\nYou want to find the roots of F(Y), i.e., those values Y* such that F(Y*)=0. \n\nSet up an iteration Yn+1 = Yn - F(Yn) / F'(Yn)\n\n\n\nwhere the derivative F'(Y)= (1/Y)+10. \n\nTry it and see if it converges for you.
669044
I'm not sure, i took that course long time ago, but I remember that the idea of solving an integrate or differential equation by series consist in expressing your function in a serie that allows an interchange with a limit (using for example dominated convergence theorem).\nThen, you will be able to derivate or integrate term by term and then after sum the resultants terms you will find your solution function .   \n\nYou should look in a book of ordinary differential equations, the topic isn't that difficult.
847056
dy/dx=y² sin x\ndy/y^2=sinxdx\n\nintegrate\n-1/y = -cosx +K\n1/y  = cosx -K\nuse initial conditions\n1=1-K\nK=0\ny= secx\n\nremember, int{sin}= -cos\n\ni hope that this helps