differential equations how to?
99672
honestly....good luck\n\ntry www.mathworld.com\n\ndo a search for differential equations, i think that would be your best bet...but that can get confusing...good luck
979217
y = sqrt(1+t^2) = (1+t^2)^(1/2)\n\nby the power and chain rule\n\ndy/dt = (1/2)(1+t^2)^(-1/2) x d/dt (1+ t^2)\n\n= 2t(1/2) /sqrt(1+t^2)\n\n= t/sqrt(1+t^2)\n\nTherefore\n\ndy = (t / sqrt(1+t^2)) dt
867402
first evaluate dy/dx = 2x +4\nnow put x=1 \ndy/dx = 2*1 + 4= 6
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
266181
dy/dx = x/y\n\nso y dy=x dx\n\nintegrate both sides u get....\n\n(y^2)/2 = (x^2)/2 + c\n\nwhere c is a constant.
881972
this is a homogeneous differential\nequation so,it should be fairly simple\n\nauxiliary equation\n\nL^2-2L-3=0\n(L-3)(L+1)=0\nL=3 or-1\n\ngeneral solution\n\ny=Ce^(3x)+De^(-x)...(1)\n\nwe now have to find C and D\n\ny'=3Ce^(3x)-De^(-x)..(2)\nnow,substitute initial values\ninto (1) and(2)\n4=C+D\n0=3C-D\n>>>>>>>C=1,D=3\n\ntherefore,particular solution\ny=e^(3x)+3e^(-x)\n\ni hope that this helps
554946
Use the equation d = sqrt((y2-y1)^2 + (x2-x1)^2)\nyou get d = sqrt((5-2)^2 + (3-(-1))^2)\n=> d = sqrt(9 + 16)\n=> d = 5 units
1001167
The solution to your question is a bit lengthy and it wont be easy typing that here. However I can tell you the guidelines to arrive at that.\n(1)Write down the equation\n(2) take differentiation wrt x on both sides.RHS will be 1 here.LHS will have to be diff..you must know how to diff e^(y/x).Use the diff rule for products.\n(3) take diff.again of the above obtained equation.you will get alot of terms then.try getting from that the form you want.
832532
dy/dx=cos x\n\nIntegrating both sides, we will get:\ny = sinx + c\n\ny(0) = sin0 + c = 0 => c=0\n\nSo y=sinx.
795982
∫_0_2 g(t) dt = 5\n\n(a)\ny= ∫_0_4 g(t/2) dt\nlet t/2 =x ; so dt = 2dx; and the range becomes x:0 to 2\nso, y= ∫_0_2 g(x) 2 dx = 2  ∫_0_2 g(x) dx = 2 * 5 = 10\n\n(b)\ny= ∫_0_2 g(2-t) dt\nlet 2-t=x; so dt = -dx; range becomes x:2 to 0\ny= - - ∫_0_2 g(x) dx = 5
741999
whats the equation or formula u need help with?
421380
from first equation put the value of b in second and then form a second degree differential equation in "a"...good luck!
589798
y' = 2x - 2\n2x - 2 = 0\n2x = 2\nx = 1
778162
Step 1: find y '\nStep 2: fill in x=1\nThat gives the right answer 2\n\nTh
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
299918
This is a separable differential equation\ndP/dt = 4P^2 + tP^2\nFactor P^2 out from the left side\ndP/dt = p^2(4+t)\nDivide each side by P^2 and multiply each side by dt\ndp/P^2 = (4+t)dt\nNow integrate each side\n-1/P = 4t + (1/2)t^2 + C\nSolve for P and you will get\nP = -1/(4t + (1/2)t^2 + C)
140101
In differential equations the Bernoulli equation is one of the form:\n\ndy/dx + P(x)*y = Q(x)*y^n\n\nwhere P(x) and Q(x) are some functions of x. Note if n=0 or n=1, the equation is linear and you can use methods of linear equations to solve it. For n>1 however, make the substitution:\n\nu=y^(1-n). This transforms your Bernoulli equation into the linear one below:\n\ndu/dx + (1-n)*P(x)*u = (1-n)*Q(x).\n\nUse the standard methods to solve this linear equation for u, and then substitute back to solve for y.