someone gud wit math plz help...int ( y+2/(2y^2+3y+1) dy )?
970256
Rule of thumb:  If the degree of the numerator is less than the degree of the denominator, you have to use partial fractions.  First, factor the denominator.  Through trial and error, you'll find that\n\n2y^2 + 3y + 1 = (2y + 1) (y + 1)\n\nFor now, we're going to ignore the integral, because what we're concerned about right now is the form of the function to integrate.  Therefore\n\n(y+2)/(2y^2+3y+1) = (y+2) /[ (2y+1)(y+1) ]\n\nThrough the knowledge of partial fractions, \n\n(y+2) / [ (2y + 1) (y+1) ] = A/(2y+1) + B/(y+1), for some constants A and B.  Our goal is to determine what A and B are.\n\nMultiplying both sides by the denominator (2y+1)(y+1), we get\n\ny + 2 = A(y+1) + B(2y+1)\n\nWe know that this is true *for all y*.  So we can solve for A and B using any value for y.  Let's choose y = -1 and then choose y=-1/2.\n\ny = -1:  \n\n-1+2 = A(-1+1) + B(2(-1)+1)\n1 = A(0) + B(-1)\n1 = -B\nTherefore, B = -1\n\ny = -1/2:\n-1/2 + 2 = A(-1/2 + 1) + B(2(-1/2) + 1)\n3/2 = A(1/2) + B(0)\n3/2 = A/2\nTherefore, A = 3\n\nSo now we know how to split up that one big fraction into two.  That is,\n\n(y+2) / [ (2y + 1) (y+1) ] = 3/(2y+1) + -1/(y+1).  So now, we integrate the new fraction.\n\nIntegral ( 3/(2y+1) + -1/(y+1) )dy = (split into two integrals and pull out constants in the process).\n\n3*Integral(1/(2y+1))dy - Integral(1/(y+1))dy =\n\n3*1/2*ln|2y+1| - ln|y+1| + C\n\n3/2 * ln|2y+1| - ln|y+1| + C\n\nReminder:  simple substitution will reveal that Integral(1/(2y+1)) = 1/2*ln|2y+1|, hence the above steps. \n\nThe above can further be simplified using rules for adding and subtracting logs, as well as the power rule for logs, but I'll leave it for you to figure out, as the more important steps were shown above.
982407
dy/dx=cosx*e^y*e^sinx\ndy/e^y= cosx*e^sinx*dx\ne^-ydy= cosx*e^sinx*dx\nintegrate\ne^-y/(-1)=int. cosx*e^sinx*dx\nput sinx=t\ncosxdx=dt\n-e^-y=int.e^tdt\n        =e^t+c\n        =e^sinx+c\nput x=0,y=0\n-1=1+c\nc=-2\n-e^-y=e^sinx-2
731860
x² - y² = 16\nSo 2x - 2ydy/dx = 0\nie dy/dx = x/y\n\nd²y/dx² = (y - xdy/dx)/y² (NOTE: x² - y² = 16  → y² = x² - 16)\n= (y - x.x/y)/(x² - 16)\n= (y² - x²)/(x²y - 16y)\n= -16/(x²y - 16y) (As x² - y² = 16)
948239
xy = y^½\n y = y^½ /x\n y = y^½*x^-½\n\ndy/dx = -½ y^½ x^ -3/2\ndy/dx = - [½ y^½ ]/x^+3/2\ndy/dx = - (y^½) / (2x^3/2)
712970
dy/dx=2xy\ndy/y=2xdx\nintegrate\nlny=x^2+C\nput in initial conditions\nln(1)=1+C\nC=0-1\n'e' each side\ny=e^(x^2-1)\n\ncheck in original equation\ndy/dx=2x*e^(x^2-1)\n2x*y=2x*e^(x^2-1)\nthis answer is correct\nat x=1.6\ny=e^(1.6^2-1)=4.758821245\n\nto apply euler's method\nto the initial value problem\ndy/dx=f(x,y),  y(x0)=y0\nproceed as follows\n(a)take x0 and Y0=y0 as starting\nvalues,choose a step size h,\nand set r=0\n(b)set x(r+1)=r+h\n(c)calculate an approximation \nY(r+1) to(y(r+1)),using the\niteration formula \nY(r+1)=Yr+hf(xr,Yr)\n(d)if furtherapproximate values\nare required,increase r by 1,and\nreturn to sub-step (b)\n\ni also have details of the \nclassical forth order runge-\nkutta method,but will keep \nthat for a later date\n\ni hope that this helps
1139627
dy/dx = 2x-3\nat X = 0. dy/dx = 2(0) -3 = -3
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
923416
Sometimes I like math.  What you need to do is add 1 to both sides of the equation for x=3y-1\nyou will be left with (x+1)=3y you should then divide the 3 from both sides... afterwards you will have (x+1)divided by 3 equals y or (X+1)/3=y\nFor the other part x+2y=9 you should substitute (x+1)/3 for the "y"\nNow solve x + 2((x+1)/3)=9\nx+(2x/3) +(1/3)=9\ntake from there
1124422
[(y^2 + y )/(y+ 2)] / [(y^2 + 3y)/(y^2-4)]\n= [(y^2 + y )/(y+ 2)] times (y^2-4)/(y^2+3y)\nFactor everything\ny^2 + y = y(y+1)\ny^2 + 3y = y(y+3)\ny^2 - 4 = (y+2)(y-2)\nCancel anything on the top with anything on the bottom\nAnswer: (y+1)(y-2)/(y+3)
718048
x + y = 3\n\nSolve for a variable.  I chose X.\nx+y = 3\n -y    -y\nx = 3 - y\n\nSubstitute the value of X in the first equation.\n2(3-y) + y = 4\nDistribute.\n6-2y+y = 4\nAdd like terms.\n6-y = 4\n-6    -6\n-y = -2\nmultiply by -1.\ny=2\n\nSince you know Y = 2, substitute that back into the second equation that you solved for X with.\nx + y = 3\nX + 2 = 3\n  -2    -2\nX = 1\n\nX = 1\nY = 2
859744
(x + 4y)(x - 4y)\nI answered 2 of your questions. Do the rest of the questions yourself.
667880
a) dy/dx = x - 1/x^2\ndy = (x-x^(-2))dx\ny = x^2/2 + 1/x + C\ny(1) = 1^2/2 + 1/1 + C = 2\n1/2 + 1 + C = 2 implies C = 1/2\nAnswer: y = x^2/2 + 1/x + 1/2, x>0\n\nb) The integration is already finished.  Just plug in -1 for x.\ny(-1) = (-1)^2/2 + 1/-1 + C = 1\n-1/2 - 1 + C = 1 implies C = 3/2\nAnswer: x^2/2 + 1/x + 3/2, x<0\n\nc) Just take the derivative of each one.  C1 and C2 are constants, so their derivative is 0.  \n\nd) C1 = 1/2 and C2 = 3/2\n\ne) Follow parts (a) and (b) except plug in 2 for x and set it equal to -1.  Similarly for y(-2) = 2.\nCheck that C1 = -7/2 and C2 = 1/2
958105
Hi,\n\nThis not just ...sqrt(pi) as one respondent thought.\n\nWhat you need are two very useful tools for integration, part one is solved by changing variables and part two is solved with integration by parts.\n\n1. let x= (y-a)/sqrt(b), so dy/dx = sqrt(b) \n=> dx = (1/sqrt(b)) dy (this isn't strictly correct but it works here)\n\nso:\nI = int(exp(-(x-a)^2/b)dx,-inf..+inf) \nI = int(exp(-((x-a)/sqrt(b))^2)dx,-inf..+inf) \nI = int(exp(-y^2)*(1/sqrt(b))dy,-inf..+inf)\nI=(1/sqrt(b))*int(exp(-y^2)dy,-inf..+inf)\nI=(1/sqrt(b))*sqrt(pi)  (from our given result)\nI = sqrt(pi/b)\n\nThe seond part requries a decent understanding of integration by parts, and if you email me I can go through this with you.
276708
2y=sqrt(1/5(y)\n10y=sqrt
861275
x(y-1) = 2y + 3\n\nxy - x = 2y + 3\nxy - 2y = x + 3\n(x - 2) = (x + 3) / y\n y = (x + 3) / (x -2)
1144370
This is a homogeneous differential equation.  \ndy/dx - y/x = x/y\ndy/dx = x/y + y/x \ndy/dx = (1/(x/y)) + y/x\nLet,\nv = y/x\ny = xv\ndy/dx = v + (x)(dv/dx)\nPlugging this back into the above equation we get,\nv + (x)(dv/dx) = 1/v + v\n (x)(dv/dx) = 1/v [canceling v out of both sides]\n1/x dx = v dv [rearranging]\nln(x) + c = 1/2 (v^2) [integrating both sides]\n2 ln(x) + c = v^2\nln(x^2) +c = (y/x)^2 [using log law and substituting value of v back into the equation]\nln(x^2) + c = y^2/x^2\ny^2 = (x^2)(ln(x^2)) + c(x^2)\nSolved!
739570
d2 - =int(c2/10)\n\neasy