wat if assume y =(x^xdx) differentiate ,tak logarithm and again differentiate to form a second degree differential equation..can that be solved?
126576
Expressions such as yours is called a power tower. They are differentiable by writing them as exponentials but they are not integrable in closed form. You can find the series form at the source link however.
20518
if y=f(x)=x^x it could be written this form: y=e^(xlnx)\nso y'=dy/dx=(1+lnx).x^x\ny'=0 => x=1/e\nif x< 1/e then y'<0 & if x>1/e y'>0\nso y have a minimun at x=1/e\nso the smallest value is f(1/e)=e^-e^-1=0.7
557507
i'm not quite sure what you mean, but if you take an ODE, say, for example, the exponential decay formula (dx/dt) + kx = 0, you can write it as dx = -kxdt, and you can solve using separation of variables.  is this what you mean?
194207
If it is a linear differential equation like\n\na(x)y'(x)+b(x)y(x)=f(x)+g(x)\nthen solve separately\naz'+bz=f\nand\naw'+bw=g\n\nThe sum z(x)+w(x) of such two "partial solutions" will give you a particular solution of your equation, as\na(x)(z+w)'(x)+b(x)(z+w)(x)\n=f(x)+g(x)\n\nP.S.\nIt seems you have not a sum of functions at the right side, but a product: (e^-t)*cost\nIf it is so, you have to see it as a unique function, and search for a particular solution of this form:\n\nx(t)=A(e^-t)cost+B(e^-t)sint\nFind the good constant A and B by taking the derivatives and placing them in the original equation.
564101
The solution to your question is a bit lengthy and it wont be easy typing that here. However I can tell you the guidelines to arrive at that.\n(1)Write down the equation\n(2) take differentiation wrt x on both sides.RHS will be 1 here.LHS will have to be diff..you must know how to diff e^(y/x).Use the diff rule for products.\n(3) take diff.again of the above obtained equation.you will get alot of terms then.try getting from that the form you want.
676900
I think I did it.  If y=1/log(x), then a few manipulations gave me\n\nx=10^(1/y).\n\nI'm assuming by log you mean log to the base 10.  I suppose mathematicians will start fussing about what happens at y=0, but a physicist like me doesn't worry that much about infinities ...
615109
Let F(a) be the value of that integral. Take the derivative with respect to a to get\nF'(a)=int x^a * log(x)/log(x) dx\n=int x^a dx\n=1/(a+1) [1^(a+1) -0^(a+1)]\n=1/(a+1).\n\nNow take the anti-derivative to get\nF(a)=log(a+1) +C.\nBut, clearly if a=0, we are integrating 0, so C=0.\nThis gives the result.\n\nThe only point at issue is whether differentiating under the integral sign is legitimate. In spite of appearances, the function inside the original integral is continuous (or can be extended to be continuous) and clearly the partial derivative is continuous in both x and a, so the result follows.
115659
It easily follows that the general solution to this differential equation is y=Ae^{kt} for some constant A [You may differentiate to double check your answers.] \n\nSo given the initial condition, i.e. y(t=60)=0.5*A, we obtain k=(ln(0.5))/(60).
348618
x = (y^2 + 2y -1) / 2\ndx=ydy+dy\n1=(y + 1 )dy/dx\ndy/dx=1/(y+1)\nd2y/dx2 = -1 / (y+1)^2
230456
well chinaren got it started but I believe his explianation is weird.\nLets assume x=3^t\nthen log3(x)+log2(x+1)=3\n=>log3(3^t)+log2(3^t+1)=3\n=>tlog3(3)+log2(3^t+1)=3\n=>log2(3^t+1)=3-t\n=>3^t+1=2^(3-t)\n=>3^t+1=2^3/2^t\n=>(3^t+1)*2^t=2^3\n=>3^t*2^t+2^t=8\n=>6^t+2^t=8\nfrom here there is one one solution possible as t=1\nso x=3^1=3
657118
Use these facts: ln(x) is the natural logarithm of x and so\nln(e^x) = x and d/dx ln(x) = 1/x\n\nTo prove d/dx e^x = e^x you find the derivative in two ways, and equate them:\n\nFirst, trivially, since ln(e^x) = x it follows that:\n\nd/dx ln(e^x) = dx/dx = 1\n\nBut also, if we let u=e^x,  then by the chain rule:\n\nd/dx ln(e^x) = d/dx ln(u) = d/du log(u)*du/dx = \n\n1/u du/dx = 1/e^x d/dx e^x (since u = e^x)\n\nEquating these 2 results we have:\n\n1/e^x d/dx e^x = 1  and multiplying through by e^x we have\n\nd/dx e^x = e^x\n\nHope that's clear enough. You are doing very well for your age. Keep up the good work!
370002
Let y = ln(ln(ln(10x)). By applying chain rule in differentiating y, we get\ny' = [ 1 / ln(ln(10x) ] * [ 1 / ln(10x) ] * [ 1 / 10x ] * [ 10 ]\n= [ 1 / ln(ln(10x) ] * [ 1 / ln(10x) ] * [ 1 / x ].\n\nBy taking natural logarithm to both sides gives\nln y' = - ln [ ln(ln(10x) ] - ln[ ln(10x) ] - ln [ x ].\n\nSince y = ln(ln(ln(10x)), the previous equation simplifies to\nln y' = - y - ln[ ln(10x) ] - ln [ x ].\n\nTaking the derivative for the both sides of the equation yields to\n[1 / y'] * y'' = - y' - { [ 1 / ln(10x) ] * [ 1 / 10x ] * 10 } - [ 1 / x ]\n      = - y' - { [ 1 / ln(10x) ] * [ 1 / x ] } - [ 1 / x ]\n      = - y' - [ 1 / x ] * { [ 1 / ln(10x) ] + 1 }.\n\nThus, \ny'' = y' * { - y' - [ 1 / x ] * { [ 1 / ln(10x) ] + 1 } }\n  = - [y']^2 - [y']*[ 1 / x ] * { [ 1 / ln(10x) ] + 1 } }, \n\nwhere y' =  [ 1 / ln(ln(10x) ] * [ 1 / ln(10x) ] * [ 1 / x ].
514572
Since you have conflicting answers, I'm just chiming in to say go with ManiacalAce -- he got the correct answer.\n\nCorrect second derivative is as follows:\n\n(-9/4)x^(-3/2) - (3/4)x^(-1/2)\n\n~ ♥ ~
497666
Take the derivative of the original function. For your given example, if I remember my calculus correctly:\nf(x)=y\nf'(x)=1\nGraph the new equation.
36538
Derivate y = A(x^3) + Bx + C to obtain:\n\ny'''=6A\ny''=6Ax\ny'=3Ax^2+B\n\n\nThen substitute in the ODE:\n\n---> 6A+12Ax-9Ax^2-3B+Ax^3+Bx+C=x\n\nArrange for x:\n\nx^3(A)+x^2(-9A)+x(12A+B-1)+(6A-3B+C)=0\n\nThe left hand side the equation above must be identically zero, so any parentheses must be zero:\n\n----> A=0 , B=1 , C=1/3\n\n---> y=x+1/3
542466
This initial value problem is surprisingly simple, since niether y nor any derivatives other than the second appear in the equation, so all that is necessary is to integrate twice:\n∫∫d²y/dx² dx dx = ∫∫6x dx dx\n∫dy/dx dx = ∫3x² + C dx\ny=x³+Cx+K\n\nNow we need merely find the constants C and K. We know that y'(0)=3(0)² + C = 0, so C=0. Further, we know that y(0) = K = 1, so K=1.\n\nThus y=x³+1\n\nb: There is exactly one such function. All antiderivatives of a function are the same up to addition of an arbitrary constant, and as we showed when solving the problem, there is only one consistent set of values for the two constants of integration introduced in this problem.
521183
dy/dx = -(e^x + e^(-x)) / -(y (e^x - e^(-x) + 1)³)\n\nSo ∫ydy = ∫(e^x + e^(-x)) dx/(e^x - e^(-x) + 1)³ \n\nLet u = e^x - e^(-x) + 1 thus du = (e^x + e^(-x) ) dx\n\nSo ∫ydy = ∫du/u³\n\nSo ½y² = -½*1/u² + ½A\n\nie y² = A - 1/u²\n= A - 1/(e^x - e^(-x) + 1)²