can you please give me little concepts(preferabbly websites) that discusses the Doolittle method in solving eq
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The sources have examples.
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just go 2 google search & write the non linear edition tutorial  \nn u will get nice tutorials
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CHeck out the following site:\n\nThe EDD just shows the flow of logic and explains the design so that everyones explanation is in the same layout.\n\nhttp://magnetweb.rockdale.k12.ga.us/French%20Partnership/students/Aaron/EDD.htm\n\nClick on the links on side in order to help you out.
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Take y on the other side\n\ny^2 dy/dx + y^3 = e^(x)\nMultiply by 3\n\n3 y^2 dy/dx + 3 y^3 = 3 e^(x)\n\nFirst term is d/dx (y^3)\nHence put z = y^3 and teh equation becomes\ndz/dx + 3z = 3 e^(x)\n\nNow this is in standard form\nMultiply throughout by e^(3x)\n\ne^(3x) dz/dx+ 3 e^(3x) z = 3 e^(4x)\nLHS can be re-written as\n\nd/dx { e^(3x)* z) = 3 e^(4x)\nIntegrate both sides\ne(3x)*z = 3 e^(4x)/4 +C\n\nz = 3/4 e^(4x)*e^(-3x) + C e^(-3x)\n\nSubstitute back for z=y^3
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this requires way more than anyone can do here, but, i know of a cool java program that will do it for you: \n\nhttp://www.math.rutgers.edu/courses/ODE/sherod/phase-local.html\n\nyou can even set your initial conditions and print it out!
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Using the abc method… \n[a]FexOy+[b]H2SO4 -->[c]Fe2(SO4)3+[d]SO2 +[e]H2O\nFe-->ax=2c------eq.1\nO-->ya+4b=12c+2d+e----eq.2\nH-->2b=2e----------eq.3\nS-->b=3c+d----------eq.4\n\nLet a=1\n\nFrom eq.1\nax=2c\nx/2=2/2c\nx/2=c-----eq.5\n\nForm eq.3\n2b=2e\n2/2b=2/2e\ne=b-----eq.6 \n\nFrom eq.2\nYa+4b=12c+2d+e\n1y+4b=12(x/2) +2d+b (Form eq.5&6)\nY+3b=6x+2d\n2d=y-6x+3b\nd=(y+3b-6x)/2------------eq.7\n\nFrom eq.4\nb=3c+d (but c=x/2)\nb=3(x/2) +d\nReplacing eq.4 by eq.7 we will get\nb=3x/2+(y+3b-6x) ½\nb= (3x+y+3b-6x) ½\n2[b=(y+3b-3x) ½]\n2b=y+3b-3x\n3x-y=b-------eq.8\n\nReplacing eq.6 by eq.8\ne=b\ne=3x-y\n\nReplacing eq.7 by eq.8\nd=(y+3b-6x)/2\nd=[y+3(3x-y)-6x]/2\nd=[y+9x-3y-6x]/2\nd= (3x-2y)/2\n\nFrom the above calculation we know that:\na=1, b=3x-y, c=x/2, d=(3x-2y)/2 & e=3x-y. \n\n[a]FexOy+[b]H2SO4 -->[c]Fe2(SO4)3+[d]SO2 +[e]H2O\nWhen replacing the values we’ll get:\n[1]FexOy+[3x-y]H2SO4 -->[x/2]Fe2(SO4)3+[(3x-2y)/2]SO2 +[3x-y]H2O
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Try this, maybe?\n\nhttp://mathforum.org/library/drmath/drmath.elem.html
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collect like terms on one side\nxdx/(1+x^2)=dy/[(1+y^2)*y]\n\n1/2* 2x/(1+x^2)= dy/y-1/2*2y/(1+y^2)\n\nBy integrating each side,\n1/2*ln(1+x^2)\n=ln|y|-1/2*ln(1+y^2)\n\nuse log rules, take square of both sides,\n\nC*y^2 / (1+y^2)=1+x^2\n\ny=sqrt([1+x^2]/[C-1-x^2])
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also try  http://mathworld.wolfram.com/ImplicitDifferentiation.html
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Try,\n\nhttp://www.personal.psu.edu/faculty/d/x/dxh40/MBLemma.pdf
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.
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use derivative rule for exponential as\n\nd/dx(A exp(bx))=(A*b)*exp(bx)\n\nyou got hint. No more Spoon Feeding
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Try:  http://www.internet4classrooms.com/eoc_algebra2.htm
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Try http://integrals.wolfram.com/
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Depending on your level of diff eq...\n\nany equation can be listed as a series expansion...and it may be sufficient (or close enough) to take the integral of "enough" terms and sum up the integrals...(the integral of the sums is the sum of the integrals).\n\nthis is a way of approximating when the higher order terms don't play a big role.\n\nThis is done in the engineering world quite a bit where the first couple of terms dominate the system (of which the diff eq is describing).  Think of a tall building (or bridge) and the wind forces upon it.  Steady winds will push on the building and make it sway but if the wind changed very rapidly, the building will sort of "dampen" out the higher order (higher frequency) inputs.\n\nAnother classic example (and one that every 4 year old has most likely mastered) is the swing.  If the person pushing the swing (or the child 'pumping his legs) does it very quickly the response of the swing won't be high, it is much more advantageous to push (or pump) at a much slower frequency (corresponding to lower orders).\n\nHopefully, this is where you are coming from...I didn't know if you wanted a rational explanation for the idea or to delve into the mathematics.
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use synthetic  divsion method \ni think this link would help\nhttp://www.purplemath.com/modules/synthdiv.htm
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The site listed below has a complete online textbook that is absolutely free and contains a well written, lucid approach to teaching this subject. \n\nHighly recommended!!\n\nhttp://www.io.com/~hcexres/textbook/