differential equation?
923150
We cannot just break it apart. dy/dx=sqrt(x/y).\nThus y*y’^2=x, and (y*y’^2)’=(x)’ that is y’^3+2*y*y’*y’’=1.\nAssume dy/dx=p, so that our task is to get parametric equations for x(p) & for y(p).\nThus p^3+2*y*p*(dp/dx)=1. Now dp/dx=(dp/dy)*(dy/dx), so that\np^3+2*y*p*(dp/dy)*p=1; (1-p^3)*dy=2*y*p^2*dp; 2*p^2/(1-p^3)*dp=dy/y.\nWhen it is broken apart: y=C*(1-p^3)^(2/3) & x=C*p^2*(1-p^3)^(2/3).\nIt’s possible, but not necessary to get rid of p & have y(x) or x(y).\nIf x=1 & y=4, then p=1/2 (& mind!!! p is not =-1/2, as p=sqrt(x/y)>=0).\nHence 4=C*(1-1/8)^(2/3), C=16/7^(2/3)
1090587
the homogeneous equation is (dy/dt) +2y = 0\nor equivalently, dy/dt = -2y\n=> (1/y)(dy/dt) = -2\nintegrate both sides wrt to t:\nintegral (1/y) dy = integral (-2) dt\n=> ln y = -2t + k   , k is a constant\n=> y = exp{-2t + k} = Ce^(-2t)  , C is another constant where C= e^k\n\n\nto solve the nonhomogeneous equation, try a particular solution of the form y = a*sin t + b*cos t\nthen, (dy/dt) = a*cos t - b*sin t\n\nsubstitute into the orginal equation to get\na*cos t - b*sin t + 2*(a*sin t + b* cos t) = - cos t\n=> (a + 2b) cos t + (2a - b) sin t = - cos t\nequating coefficients:\na+2b = -1 and 2a-b = 0\nsolve for a and b:\na = -1/5, b = -2/5\n\nhence, a particular solution to the nonhomogeneous equation is y= -1/5 sin t - 2/5 cos t\n\nso, the complete general solution to the nonhomogeneous equation is\ny = C e^(-2t) - 1/5 sin t - 2/5 cos t\nwhere C has to be determined from initial conditions
697597
1. Solving for dy/dx gives dy/dx = -y/(x+y), but this doesn't fit the form of the Bernoulli equation.  Instead, flip it around and get dx/dy + (1/y)x = -1, so n=0, P(y)=1/y, and Q(y)=-1.  In this case, there's no need for a change of variable, so instead just jump directly to using the integrating factor m(y) = e^(integral of 1/y) = y, to get\n\ny dx/dy + x = -y, so (xy)' = -y.  Therefore\n\nx = (1/y) (-y^2/2 + C), or x = -y/2 + C/y.\n\nMultiply through by 2y: 2xy + y^2 = 2C, which is equivalent to the solution you expect, since C is an arbitrary constant.\n\n2. Since n=2, let v = y^(1-n) = 1/y.  After transforming, you get \n\ndv/dx + vp(x) = q(x)\n\nwhere p(x) = (1-n)P(x) = 1/x and q(x) = (1-n)Q(x) = x, so you're solving\n\ndv/dx + (1/x)v = x.\n\nNow use the integrating factor x, and  you have\n\nx dv/dx + v = x^2\n\nso (vx)' = x^2, or v = (1/x)(x^3/3 + C), so\n\nv = x^2 / 3 + C/x.  But v = 1/y, so \n\ny = 1 / (x^2/3 + C/x) = x / (x^3/3 + C).\n\nThis doesn't agree with what you're saying the answer should be, but it does satisfy the differential equation (as I confirmed by plugging in this solution), whereas your answer doesn't.
934968
Okay, assuming you mean:\n∫e^(2x)/√(1-e^(2x)) dx\nLet u=1-e^(2x), du=-2e^(2x) dx, dx = -1/2 e^(-2x)\n-1/2 ∫1/√u du\n-1/2 ∫u^(-1/2) du\n-u^(1/2) + C\n-√(1-e^(2x)) +C
830837
If ∫f(x)dx = y(x), then f(x) = d y(x) / dx;  So to get the answer, take the derivative of the expression x^2 - 2x + 1.  The value of the integral at limit 1 is  a constant and will not appear in the derivative.
803550
With respect to what?\n\nd/dx xy = y + x dy/dx\n\nd/dy xy = x + y dx/dy\n\nd/dt xy = x dy/dt + y dx/dt\n\nIt's just the Product Rule.
1068054
well Y' = dy/dx\nso using seperation of veriables we can get\ndy/dx = K - 1/3*y^3\nand dy/(K-1/3*y^3)=dx now integrate both sides. \nDoes that help?
279194
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).
940656
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
106013
y = -2x + 8\n\nsolve it yourself
904469
Let u = e^x + 1.  Then du= ...\n\nHope that helps.
314068
y = x^e^(7x)\nln y = ln [x^e^(7x)]\nln y = e^(7x)*ln x\n(1/y)*y' = 7*e^(7x)*ln x + (1/x)*e^(7x)\n(1/y)*y' = e^(7x)*[7*ln x + (1/x)]\ny' = y*[e^(7x)*[7*ln x + (1/x)]]\ny' = x^e^(7x)*e^(7x)*[7*ln x + (1/x)]\n\nTo ingameswetrust:\nHow could y = x^e^(7x) be mistaken for a differential equation? There are no derivatives anywhere in the equation. The author clearly states, "find d/dx..." However, if you want to solve a differential equation, then solve the following:\ndy/dx = x^e^(7x)*e^(7x)*[7*ln x + (1/x)]\n\nThere is one answer that should be obvious even to the casual observer.
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
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
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
785428
y=e^cos(x) + cos(e^x)\ny'=-sin(x)e^cos(x) + -sin(e^x)e^x\n=-(sin(x)e^cos(x)+sin(e^x)e^x)
914813
Your correct, it is equal to f(x).\n\nthe integral and the differential cancel eachother out.!