Consider the differential equation \n(E): y”+ 2y’ + y = -2 e to the power of -x                                \nLet y= ze-x \na)Form the differential equation (F) satisfied by z\nb)Solve (F) and deduce the general formula of (E)
210567
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)
328954
(1) { -2x-y-z=0\n(2) { -x-2y-z=0\n(3) { -x-y-2z=0\n\nLet's look at (2):\n-x-2y-z=0 => x=-2y-z\nNow at (3):\n-x-y-2z=0 => x=-y-2z\nNow we can see that\n-2y-z=-y-2z => *** y=z ***\n\nNow let's put these values into (1):\n-2x-y-z=0 => -2x-y-y=0 => -2x-2y=0 => 2x=-2y => *** x=-y ***\n\nFor now, we know that x=-y=-z\nLet's put all of what we know into (2), and express everything using 'x':\n-x-2y-z=0 => -x-(-2x)-(-x)=0 => -x+2x+x=0 => 2x=0 => *** x=0 ***\n\nThus, x=0 => y=-0=0 => z=0\nSo the only solution is (0,0,0).\n\nGood luck!
419066
1.\nc=(xy^2-1)/y\ndifrentiate w.r.t x\n(y(y^2 + 2xyp)-p(xy^2-1))/y^2=0\ny^3 + 2xy^2p - pxy^2 + p=0\ny^3 +xy^2p +p=0\nwhere p=(dy/dx)\n2.\nsimilarly diffrentiate \n(x+y^2)/x^2=-c\nto get\n2(x^2)yp - x^2 - 2xy^2=0\nthanks
442548
okay well\n\n1) combine two to figure out the first variable\nx-2y+z=4\n-x+2y+3z=-4\nso 4z=0; z=0\nthen figure out the next variable with a different set of equations\nlike and multiply so the ys will cancel [take Zs out cause zero]\nx-2y+z=4\n2(2x+y)=6\n\ntherefore\nx-2y=4\n4x+2y=6\n5x=10\nx=2\n\nthen substitute for y\n\n-x+2y=-4\n-2+2y=-4\n2y=-2\ny=-1\n\nso y=-1, x=2, z=0\n\n\ncontinue the process with the rest.
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(6y^2-3y)(y+7)-->(2y)(y-1)(y+7)\nif that helps
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-2x+2y+2z=-22\ndividing by -2\nx-y-z=11\nmultiplying by 2\n2x-2y-2z=22\n-2x+3y+z=-18\nadding\ny-z=4\nsubtracting (2) from (1)\n-y+z=-4thissysem cannot be solved\nyou may have to guess check and revise
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2x+2y+3z=-1 .....(1)\n3x-1y-2z=18 .......(2)\n-3x+1y-1z=-9 ......(3)\n\n(1)+(2)x2\n2x+2y+3z=-1\n6x-2y-4z=36\n-------------------+\n8x-z=35 .............(4)\n\n(2)+(3)\n3x-1y-2z=18\n-3x+1y-1z=-9\n---------------------+\n-3z=9\nz=-3\n\nsubstitute this value of z to (4)\n8x-z=35\n8x-(-3)=35\n8x+3=35\n8x=32\nx=4\n\nsubstitute the value of x and z to (1)\n2x+2y+3z=-1\n2(4)+2y+3(-3)=-1\n8+2y-9=-1\n2y-1=-1\n2y=0\ny=0\n\n(x,y,z)=(4,0,-3)\n\nhope this is clear enough
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Confirmed!\n\ndy/dx = 4(3+x)^3/(1+x)^4 - 4(3+x)^4/(1+x)^5 (I assume you know how to get to this, since you got the right answer.\nAt x=1, dy/dx = 4*4^3/2^4 - 4*4^4/2^5 = -16\nand y = (4/2)^4 = 16\nUsing the point-slope form:\ny-y1 = m (x-x1)\ny - 16 = -16 (x - 1)\ny=-16(x-1) + 16   (the form you used)\ny= -16 x + 32
450839
(E): y+2=-3/2(x-6) (=) y=-3/2*x + 7
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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.
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Multiplying eqn no 1 by 3 and subtracting it from eqn.no2,w get \ny-2z=5........(eqn.no4)\nmultiplying equn. no.1 by 2 and adding it with eqn. no.3,we get\n3y-5z=6.......(eqn. no 5)\nmultiplying eqn, no(.4)by 3 and subtracting it from eqn. no.(5),w get  z= -9\nPutting the value of z in eqn. (4),we get\ny+18=5\nTherefore  y=5-18= -13\nPutting the value of y and z in equation no.1,we get\nx-26+9=3\n=>x=3+26-9=20\nTherefore,x=20,y=5 and z= -9   ans.
34661
dy/dx = (4x^2 - 2xy)/(x^2 + 1)\nor, (4x^2 - 2xy)dx = (x^2 + 1)dy\nor, (4x^2 - 2xy)dx - (x^2 + 1)dy = 0\n\nlet \nM = (4x^2 - 2xy)\nN = -(x^2 + 1)\n\nδM/δy = -2x  &\nδN/δx = -2x\n\nHere,\nδM/δy = δN/δx; hence it is an exact equation\n\nNow,\n∫Mdx = 4(x^3)/3 - y(x^2)  --- (1)\n∫Ndy = -y(x^2) - y        --- (2)\nRejecting the tarm (-y(x^2)) in (2) which is already occers in (1) and then adding (1) & (2) and equating the sum to a constant , we get the general solution to be\n4(x^3)/3 - y(x^2) - y = C ---(solution)\n----------------------------------------------------------
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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).
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Well...\nIf Y/Y'=k/y then Y'/Y=y/k\nso (lnY)'=y/k\nIntegrating\nlnY=ln A+ (y^2)/2k      A- constant\nExponentiating\nY=A*exp( (y^2)/2k )\nAm I right?
439219
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???
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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
28187
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