hi how  solve this defrintioal equption  { x^2*y"-x*y' +4*x*e^(x^3)=0} thank you.?
492313
x^2*y"-x*y' +4*x*e^(x^3)=0\n\nx[x*y"-y' +4*e^(x^3)]=0\n\n=> x=0 or x*y"-y' +4*e^(x^3)=0\n\nx*y"-y' +4*e^(x^3)=0\n\n=>\n\n∫[x*y"-y' +4*e^(x^3)].dx=0\n\n=> [-4*x*Gamma[1/3, -(x^3)]]/[3(-(x^3))^(1/3)]=0\n\nwell even my computer couldn't solve that, so let the answer be x=0.
925629
polyeig(-5,-2,4,2) \n\nwill solve the equasion.  To learn about interpelations type:\n\nhelp interp1
965203
y = (ln x²)^(2x+3) = e^((2x+3) ln ln x²)\nNow just ruthlessly apply the chain rule:\ndy/dx = e^((2x+3) ln ln x²) * (2 ln ln x² + (2x+3) * 1/ln x² * 1/x² * 2x)\nAnd simplify:\n(2 ln x)^(2x+3) * (2 ln (2 ln x) + (2x+3)/(x ln x))
900593
y=f(x)= a*x*e^(b*(x^2)),\n y’=a*[e^(b*(x^2))+x*b*2*x*e^(b*(x^2))],\n y’=a*[1+2*b*x^2]*e^(b*(x^2))=0,\nThe value of e^(b*(x^2)) is always >0, also ‘a’ is not =0. \nthen 1+2*b*x^2=0, while x=3, b=-1/(2*3*3)=-1/18\n& 4=a*3*e^(-3*3/18), hence a=(4/3)*sqrt(e)\nis it really MAX?\ny’(0)=a>0 means increasing function, thus YES!
949784
y= (x^2 - 1)^2\nx^2-1=+/-√y\nx^2=1+/-√y\nx=+/-√(1+/-√y)
27685
You have two equasions 2x=3y and y=2x+4\nYou're going to plug one equasion into the other; plug the y= into the  first equasion because the y is already by itself.\n2x=3(2x+4) (I plugged what y equaled into the y in the first equasion.\nThen you solve for x\n2x=3(2x+4)\n2x=6x+12\n-4x=12\nx=-3\nNow that you have solved for x, plug x back into the y='s equasion\ny=2x+4\ny=2(-3)+4\ny=-6+4\ny=-2
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
1131726
This is a First Order Differential Equation to solve it:\n\nx y' - ay = b x^4\n\nRewrite it by dividing by x:\n\ny' - (a/x) y = b x^3\n\nwhere P(x) = -a/x  ....   and q(x) = b x^3\n\nNow find the integration factor which is \n\nu(x) = exp ( Int(P(x) dx) ...... where P(x) = (a/x)\n\nu(x) = exp ( Int (-a/x) dx ) \nu(x) = exp( -a Ln(x) = x^(-a) = 1/(x^a)\n\nNow the solution of the DE is\n\nu(x) y = Int ( u(x) * q(x) )\n\n1/(x^a) * y = Int ( 1/(x^a) * b x^3)\n\n1/(x^a) * y = Int ( b x^ (3-a) )\n\n1/(x^a) * y = [b / (4-a) ] x ^ (4-a) + C\n\ny = b /(4-a) x ^ 4 + C x^a
830587
y'=-x^(-1/2)\ny"=(-1/2)x^(-3/2)\n2x^2[(-1/2)x^(-3/2)]+3x[x^(-1/2)]-x^(-1)=0\n-x^(1/2)+3x^(1/2)-1/x=0\n2x^(1/2)-1/x=0\n2x(x^(1/2))-1=0\nx^(3/2)=-1/2\nx=(-1/2)^(2/3)\n\n\ni think thats correct... but who knows...>.<;;;\nis there an answer key???
1109888
Jennifer, I'm a little bit puzzled about your question because you're asking for what the funtion is and you have it written there.  The funtion is s = -16t2 + v0t + s0\n     What I really think you want is the solution.  Correct?  Well, here's how you get that.  You put in 0 for s0, and 64 for v0.  ALSO, s, on the left side of the equation, is also zero because it will land back on the ground when it's done.  When you solve using these numbers, the only variable left is t.  And when you do the algebra, \n  0  = -16t2  +  64t  + 0\n \n16t2  =  64t\n\n16t  =  64\n\nt  = 4      seconds, that is.\n\nI hope this works.  I'm a physics teacher.  If you need further clarification for this problem, or if you need help with any others, feel free to write me at fortitudinousskeptic@yahoo.com\nI'll be glad to help!
465353
You must have meant y = z e^(-x). Substituting: \n\ny' = (z'-z)e^(-x)\ny'' = (z''-2z' + z)e^(-x)\n\ninto the equation gives (answer to (a)):\n\nz'' = -2\n\nwhich has solution (answer to (b)):\n\nz = -x^2 + Ax + B\n\nwhere A and B are constants, so the solution to the original equation is:\n\ny = (-x^2 + Ax + B)e^(-x)
973940
A previous answer was incorrect. It is\n\n-4x(x+3y)(x-2y)
703729
You want to start by integrating the equation. Fortunately, this is not difficult.\n\ndy/dx = (x-2)e^(-2y)\n\nInt(e^(2y)dy) = Int((x-2)dx)\n\n (1/2)e^(2y) = (1/2)x^2 - 2x + C, where C is some constant\n\ne^(2y) = x^2 - 4x + C  Note that e^(-2y) = 1/[x^2 - 4x + C]\n\n2y = ln (x^2 - 4x + C)\n\ny = (1/2)ln (x^2 - 4x + C)\n\nWe are almost done, now we just need to plug in the given condition, and solve for C.\n\nln (2) = (1/2)ln(4 - 8 + C)\n2ln(2) = ln(C-4)\n4 = C - 4\nC = 8\n\nSo the particular solution is y = (1/2)ln (x^2 - 4x + 8)\n\nIf you want to check this, we need to differentiate it\n\ndy/dx = (1/2)(2x-4)/(x^2 - 4x + 8)\n\nRemember from above that  e^(-2y) = 1/[x^2 - 4x + 8]\n\nSo dy/dx = (x-2)e^(-2y) as required
333390
S (x^2)(e^2(x^3)) dx\n\nlet u = x^3\nthen du = 3x^2 du\n\nS (x^2)(e^2(x^3)) dx = 1/2 S e^2u du\n= 1/3[(1/2)e^2u] + C\n= (1/6)e^(2x^3) + C\n\nAnswer C.
821550
if f(x) = ∫[t = x to 0] 1/ (1 + t + t²) dt\n\nf'(x) = -1/ (1 + x + x²)\n\nf''(x) = (1 + 2x)/(1 + x + x²)²\n\nFor concave UP f''(x) > 0\n\nSo 1 + 2x > 0 (as (1 + x + x²)² > 0 for all x)\n\nie x < -1/2
835686
4x^2 + 3y^2 = 4\n8x x' + 6y y' = 0\n8 x '' + 6 y'' = 0\n6y'' = -8x ''\ny '' = -4x / 3
1057515
Please check the statement of the problem, then apply this.\n\nWe know that\nsin (x)=(e^(ix)- e^(-ix))/2i\n\ncos (x)=(e^(ix)+ e^(-ix))/2\n\nalso\n\nsin(ix)=i(sinh(x))=(e^(-x)- e^(ix))/2i\n\ncos(ix)=(cosh(x))=(e^(-x)- e^(ix))/2\n\nThe only thing you need is if z=x+iy then |z|=sqrt(x^2 + y^2)\n\nGood luck