who can solve this differential equation? dx=(me^(-ky) - x)dy?
822532
This is an interesting problem, because we are solving for x instead of y (given the equation).  First divide by dy, thus you get:\n\ndx/dy = me^(-ky) - x \n\nWe cannot use separation of variables for this O.D.E. because it is not of the form dx/dy = g(y)h(x).  Thus, the easiest way is to utilize an integrating factor.  Rewrite the equation as:\n\ndx/dy + x = me^(-ky)\n\nNow multiply through by e^y:\n\ne^y(dx/dy) + xe^y = me^(-ky)e^y = me^(-2ky)\n\nWhen utilizing an integrating factor, you try to make one side look like the product rule of the derivative.  By inspection, you can see that the LHS of this equation is:\n\nd[xe^y]/dy\n\nWe can double-check this:\n\nd[xe^y]/dy = e^y(dx/dy) + x(d[e^y]/dy) = e^y(dx/dy) + xe^y\n\nTherefore, our equation simplifies to:\n\nd[xe^y]/dy = me^(-2ky).  Now taking the integral of both sides:\n\nxe^y = [-1/(2k)]me^(-2ky) + C\n\nDividin by e^y and get your solution:\n\nx = (-m/[2k])e^(-3ky) + C\n\nIf you are looking for y, just take the inverse of this equation (interchange x and y and solve for x).\n\n-----------\n\nHope this helps
1055222
Is that e^y?\n\nCompute the differential of both sides\n\n7dxe^y + 7xe^ydy + 9x^2dx = 20y^3dy\n(20y^3 - 7xe^y)dy = (7e^y + 9x^2)dx\ndy/dx = (7e^y + 9x^2)/(20y^3 - 7xe^y)
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
982897
the 1st one is simple\n(x+y) dy/dx + y =0\nor xdy/dx + ydy/dx + y =0\nor (xdy/dx + y)  + ydy/dx = 0\nxdy/dx + y = d/dx(xy)\n\nand d(y^2/2)/dx \n\nso d/dx(xy+y^2/2) = 0\n\nor xy+y^2/2 = c\nor 2xy+y^2 = 2c or C\n\nproved\nfor the second one\n\ndy/dx -(1/x)y = xy^2\n\nlet y/x = t\ny = xt\ndy/dx -y/x = t+ x dt/dx -t = x dt/dx\nso\nx dt/dx = - x (xt)^2\nor dt/dx = x^2t^2\ndt/t^2 = x^2 dx\nyou should be able to proceed
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
719380
Let y = x^(1/2)\n\nso y^2 = x,\n\ndx = 2y dy\n\nThe problem becomes:\n\nint [1 / (y + y^3)] 2y dy\n\nOr:\n\nint 2 [1 / (1 + y^2)] dy\n\nA recognizable integral. Integrate and substitute back x^1/2 = y
92557
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
304526
let a-x=u\nso, -dx=du\n\nthen ∫1/(a-x)^2dx= \n-∫1/[(u)^2]du=\n-∫u^(-2)du=\n\n-[u^(-2+1)]/(-2+1)=\n\n-u^(-1)/-1=\nu^(-1)=\n\n1/u + C=\n\n1/(a-x) + C
765864
I'm unclear with the way you stated the problem. Am I looking for a implicit derivative. I'll check back to solve it.
757602
[(x^3)/(2 - x)]^4/3 \ndy/dx = (4/3)[(x^3)/(2 - x)]^1/3[3x^2(2 - x) + (x^3)]/(2-x)^2\ndy/dx = (4)[(x^3)/(2 - x)]^1/3[6x^2 - 3x^3 + (x^3)]/(3(2-x)^2)\ndy/dx = 8x^3(3 - x)/(3(2-x)^7/3)
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.
1166733
dy/dx = {(3x^2-2)(4x-3)^4 - (x^3-2x)(4)(4x-3)^3(4)}/(4x-3)^8\n= {(3x^2-2)(4x-3)^4 - 16(x^3-2x)(4x-3)^3}/(4x-3)^8\n\nUsing the quotient rule.  d/dx f(x)/g(x) = {f'(x)g(x) - g'(x)f(x)}/g(x)^2
1034892
What is the function? I assume it is \n\n(dy/dx is equal to 1 only if y=x+C whre C is a constant)\n\nTo prove use the definition of the derivative which is dy/dx as a limit of delta_x appraching of (delta_y/delta_x)\nwe have \ny2-y1=x2+C-x1-C then\ndelta_y=x2-X1\n\ndelta_y/delta_x=(x2-x1)/(x2-x1)=1 QED\n(the is no need to take a limit, however in must be used. I hav eommited it for lack of better representation)\n\nBest
69673
The Question is reversed \n\ny =4e^(-x) sin3x     ---------    (eqn1) \n\ny'=[4e^(-x) 3cos3x] + [-4e^(-x) sin3x]\n\ny'=[4e^(-x) 3cos3x] - [y] ----- From(eqn1)y =4e^(-x).sin3x \n\ny'=[4e^(-x) 3cos3x] - [y] ------(eqn2)\n\ny"=[4e^(-x)(-9)sin3x] + [ -4e^(-x)3cos3x ] - y' \n\ny"+y'=[-9y] - [4e^(-x) 3cos3x]   ----From(eqn1)y=4e^(-x)sin3x \n\ny"+y'=[-9y]- [y'+y] ------ From(eqn 2) 4e^(-x).3cos3x =y'+y\n\ny"+2y'+10y=0\n\nSo this differential equation has  y=4e^(-x)sin3x as one of it's solutions
980558
∫xe^x dx \n= ∫x d(e^x)/dx dx \n= xe^x - ∫e^x (dx/dx) dx\n= xe^x - ∫e^x dx\n= xe^x - e^x + c\n= (x - 1)e^x + c
1139073
Differential Equations
899944
first, I would put it in an easier to express format g(t)=[t^(1/2)](1-t)  then take the derivative g'(t)=[(1/2)t^(-1/2)](1-t)-t^(1/2)   now just solve for g'(t)=0, so 0=[(1/2)t^(-1/2)](1-t)-t^(1/2)      now we distribute to get 0=[(1/2)t^(-1/2)]-(3/2)t^(1/2)     we add (3/2)t^(1/2) to both sides to get (3/2)t^(1/2)=(1/2)t^(-1/2)   simplfy this to [3sqrt(t)/2]=[1/2sqrt(t)]   we multiply both sides by 2sqrt(t) to get 6t/2=1 or 3t=1, which means that t=1/3\n\nAnd by the way, this isn't the chain rule, this is the Product Rule