Which should i take first? Calculus 3 or Differential Equations?
349036
If Calculus 3 implies that you have coursed through Calculi 2 and 1, then Differential Equations would be a meaningful course.\n\nDifferential Equations are the bread and butter of engineering, of Building anything; while Calculus 3,4, etc would enable you to find patterns, study patterns, build theories (in general, play with data). Differential Equations would enable you to construct problems and then solve them.\n\nSo, I guess it depends, as always :-), on your goals (both immediate and future).
1068054
well Y' = dy/dx\nso using seperation of veriables we can get\ndy/dx = K - 1/3*y^3\nand dy/(K-1/3*y^3)=dx now integrate both sides. \nDoes that help?
887510
Differential equations.\nLinear algebra.\nNumerical analysis.
1091033
hardly calculus, simple algebra\n\n\n1/3 = B / (120 +2B)\n40 + 2B/3 = B\n40 = B/3\nB=120
886519
The first one is non-linear (because of the (y')^2 term) and I am not an expert on non-linear differential equations.  Are you sure that is right because if it is only y', then you can simply seperate variables and then solve it.\n\nThe second one, find the integrating factor and then it becomes an exact equation.
235488
the best is you get a good book about initial value problems... go und get one at your closest library...\n1.- A First Course in Partial Differential Equations With Complex Variables and Transform Methods by Hans F Weinberger\n2.- A First Course In Differential Equations With Modeling Applications by Dennis G Zill\n3.- Calculus: Concepts and Contexts (with Tools for Enriching Calculus, Interactive Video Skillbuild... by James Stewart\n----------------------\nmany sites haves definitions about the topic:\nhttp://mathworld.wolfram.com/InitialValueProblem.html\nhttp://planetmath.org/encyclopedia/InitialValueProblem.html\nhttp://laps.fsl.noaa.gov/~jedwards/thesis/sectionA_2.html
744437
The most fundamental in calculus is the concept of limits, as defined, calculus is either differential or integration, differential being the breaking up into smaller parts while integral is the reverse, both needs the limits in their operation.
117809
Now you're getting into Calc III with binormal planes, curls, divergents, and such.  There is a Hell-ish formula somewhere.
881972
this is a homogeneous differential\nequation so,it should be fairly simple\n\nauxiliary equation\n\nL^2-2L-3=0\n(L-3)(L+1)=0\nL=3 or-1\n\ngeneral solution\n\ny=Ce^(3x)+De^(-x)...(1)\n\nwe now have to find C and D\n\ny'=3Ce^(3x)-De^(-x)..(2)\nnow,substitute initial values\ninto (1) and(2)\n4=C+D\n0=3C-D\n>>>>>>>C=1,D=3\n\ntherefore,particular solution\ny=e^(3x)+3e^(-x)\n\ni hope that this helps
650498
In fact, the calculus is simple but if you have problems you need to take many classes before it. However, if you have no math clue...well you should start again since the beginning. You can do it, ask for help always and study together with your classmates.
419729
Why yes they are, you are just like me (just younger), i took calculus 1, 2 @16, Calculus 3 and differential equations @17, Vectors and matrices, numerical analysis, and probability and statistics, Partial differential equations 1 @ 18... you get the picture, if you keep up with the math, you can graduate out of highschool with a major in math.  That is a great gift you can give yourself.  I applaud that a 12 year old is so interested in math.  Keep up the awesome work.
262508
Differential and Integral Calculus\nTopology of the real line\nLinear algebra\nAbstract algebra\nComplex Analysis\nDifferential Equations
117843
Sure.  The derivative of x^3 is 3x^2.  That should get you started.
1043256
As you can tell from the various answers, your question is ambiguous.  If y is a function of x, then d(y^3)/dx = 3y²(dy/dx), using the chain rule, not + dy/dx as metaraison would have it.  If y is not a function of x, then dy/dx = 0 as if y were a constant.
701308
no.\nif u=tan3x you take the derivative of the function:\n(sec 3x)^2 and then multiply it by the derivative of what's INSIDE the function (the chain rule). THe derivative of 3x is 3.\ndu= 3(sec 3x)^2.\n\nHere's the chain rule explanation\nif u=f(g(x)) then du= f'(g(x)) * g'(x)\nin words:\ndu is f prime of the original function multiplied by the deriv of what's inside the fuction.
48881
Anytime you have a problem that is continuous, as opposed to discrete, you will need to use either integration or differentiation to characterize the solution.  That means to say that every problem that involves time in one way or another will necessitate the use of calculus to find an answer at some instant in time.  If the problem is such that you can approximate the answer, then you can discretize the parameters as small slices of time and come up with an approximate answer - without using calculus.
965952
Implicit differentiation is really jsut an exercise in using the chain rule. For instance, if you had e^(x^2), its derivative would be e^(x^2) * 2x, since d(x^2)/dx = 2x. It's the same thing with implicit, except that you don't actually know what dy/dx is yet, so you just write dy/dx. So for my example, d(e^y)/dx = e^y dy/dx. Now, onto your problem:\n\nd(x/(x+y))/dx = ((x+y) - (1+dy/dx)x)/(x+y)², or being simplified, (y-x dy/dx)/(x+y)²\n\nOf course, this assumes you are differentiating with respect to x. If you are differentiating w.r.t. a third variable (e.g. t), then you would have to use the chain rule on x as well. So your derivative would be:\n\nd(x/(x+y))/dt = (dx/dt (x+y) - (dx/dt + dy/dt)x)/(x+y)²