Does anyone know first order differential equations?
439816
set up the diffeq:\n\ndM/dt=1.05*M+alpha //money accrued at rate of 5.5% per year plus some amount alpha that is placed into the accout per year\ndm/dt=1.05*m -45000 //rate spent per year after retirement (assuming interest continues to accrue)..\nM(0)=162000; //intial ammount in retirement fund\nM(15)=m(0) //retirement fund begins expenditure with intial value of total saved + interest, or M(15)...\n\n// Note: 12 is measured from date of retirement...\nm(12)=0; //money only lasts 12 years, assume total amount expended...\nsolve dm/dt and plug in m(12)=0 to figure out what the total saved, m(0), must be present at the beginning of retirement in order to completely expend the money in the 12 yr period\n\nfinally, solve for M(t) and use the known values at M(0) and M(15) to compute the value of alpha.
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)
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
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?
184556
Let x(t) = A*cos( w*t + theta )\n\nthen x(0) = A*cos( theta )\nand dx/dt (0) = -A*w*sin(theta)\n\nso if you know x and its derivative in t=0 u have two equations and two unknowns ( A and theta ) . the rest is simple algebra .
341524
Second order polynomial equation in a single variable is called quadratic equation.\nSo the answer is no.
1157760
I think you are saying this:\n\nN'(t)=A*N(t*k)  \n\nWhere A, k are constants and N'(t) represents the differential, or derivative.\n\nProblem:  the constant on the inside of function N on the right hand side.  If it wasn't there, this would be a nice differential equation you could solve by using introductory DE methods.  I'll assume your tilda (or my "k") isn't there.\n\nJust by looking at the problem, you can tell that the original function N(t) is an exponential form.  \n\nN'=A*N\nN'/N=A\nd/dt(ln(N))=A\nint(d/dt(ln(N)*dt)=int(A*dt)\nln(N)=A*t + C\ne^ln(N)=e^(A*t+C)\nN(t)=(e^(A*t))*e^C\ne^C is just a constant, I'll rename D\nN(t)=D*e^(A*t)\n\nSolved.  For future reference, learn to enter a problem properly on this forum.  We can't read your mind.
761744
use the product rule, with the first term as t and the second as (4-t)^1/2\nproduct rule says that the derivative is second term times the derivative of the first term plus first term times the derivative of the second term.  Don't forget to use the chain rule along the way.\n\nThat is\nxy would have a derivate of yx' =xy'\n\nSo here\n[(4-t)^1/2](1) + (t){[1/2(4-t)^ -1/2][-1]\n(4-t)^1/2 + -t/2(4-t)^ -1/2
217134
Hi there,\nThis is a linear differential equation and can be solved using a certain list of steps (check any calculus text book). To avoid finding an integrating factor etc, we note that by dividing by x^2 we get:\nf'(x)/x - f(x)/ x^2 = 1/x\nThe left side is the same as (f(x)/x)' [product rule].\nThus we have, (f(x)/x)' = 1/x\nIntegrating gives f(x)/x = ln|x|+C\nSo, f(x)=x ln|x|+Cx\nSubstituting f(-1)=1 gives C=-1\nThus, f(e^-1)=(e^-1) ln|e^-1| - e^-1=-2e^-1.\n\nCheerios,\nMike
873172
u can get them with the help of second and higher derivatives.\ninflection point is there where the second derivative is zero.
292019
If the unit vector has components "a" and "b" (the unknowns for which you want to solve) then your first equation is\n\n1.  a^2 + b^2 = 1\n\nIf you have f and the directional derivative then you can take the partial derivatives of f with respect to x and y (call these f_x and f_y) and get your second equation\n\n2.  a f_x + b f_y = directional derivative\n\nWith two equations in the two unknowns a and b you then try to solve for a and b.
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
1178962
To solve this it's a pretty straightforward application of Laplace transforms. You can do this question using convolutions as follows:\n\nFirst take the laplace transfom of the ODE:\n\nlet L{y(t)}(s) = Y(s) be the laplace transform of y(t).  Then we have:\ns^2(Y(s))  - 2s + 3Y(s) = 1/s^2 - (e^-s)/s\n\nrearranging we have Y(s) = (1/s^2 - e^-s/s + 2s) (1/(s^2+3)) = G(s) H(s) where G(s) = (1/s^2 - (e^-s)/s)  and H(s) =  (1/(s^2+3)) \n\nnow using the inverse laplace formula for the convolution of two functions:\n\ny(t) = L(^-1){Y(s)} = integral from 0 to t of (h(t-x)g(x)dx) + 2cos(sqrt(3)t)\n\n where h(t) = 1/sqrt(3) sin(sqrt(3)t) and g(t) = (t - u1(t))\n\nand the integral can be evaluated. \n\nSorry for the readability but I don't think yahoo answers can render math text properly. Also, as a note, this question would be far easier to solve using a partial fraction method rather than unecesarily using convolutions.
1111738
the best way to appr this problem is to apply the first principle i.e.\nf'(x)= Lt     [f(x+h)-f(x)]/h\n       h->0\nnow putting f(x)=e^x we have\nf'(x)= Lt    [e^(x+h)-e^x]/h\n       h->0\n     =Lt    [e^x(e^h-1)]/h\n       h->0\n\nnow (e^h-1)/h where h tends to 0 is 1\nhence f'(x) = e^x\ntherefore the derivative of e^x is e^x
660658
Let y = f(x) be the solution with y(0) = 1, y'(0) = 0.\n\nLet y = g(x) be the solution with y(0) = 0, y'(0) = 1.\n\nThen the solution for y(0) = c0, y(1) = c1 is just:\ny(x) = c0 f(x) + c1 g(x)\n\nAs power series, we can write:\n\nf(x) = 1 + 0 x + a2 x^2 + a3 x^3 + a4 x^4 + ...\n\ng(x) = 0 + 1 x + b2 x^2 + b3 x^3 + b4 x^4 + ...\n\nThen f''(x) = 2 a2 + 6 a3 x + 12 a4 x^2 + ...\nwith (n+2) (n+1) a{n} as the coefficient of x^n.\n\nLikweise for g''(x) but replacing the a's with b's.\n\nFinally, sin(x) = x - (1/3!)x^3 + (1/5!)x^5 - ...\nwith all x^even coefficients equal zero, and\nwith the x^(2k+1) coefficient equal to (-1)^k / (2k+1)!.\n\nNow just express both\n\nxf'' + sin(x) f = 0\nxg'' + sin(x) g = 0\n\nas power series on the left (messy) and right (all zeroes)\nand equate left and right (zero) coefficients for each of\nx^0, x^1, x^2, x^3, etc.\n\nFor f you have\n\n(2 a2 x + 6 a3 x^2 + 12 a4 x^3 + ...) +\n (x - (1/3!)x^3 + (1/5!)x^5 - ...) (1 + x + a2 x^2 + a3 x^3 + ...)\n = 0\n\n0 x^0\n + (2 a2 + 1) x^1\n + (6 a3 + 1) x^2\n + (12 a4 + a2 - (1/6)) x^3\n + ...\n = 0 = 0 + 0 x + 0 x^2 + 0 x^3 + ...\n\nEquating coefficients, you get\n\n0 = 0\n that is good!\n\n2 a2 + 1 = 0\n or a2 = -1/2\n\n6 a3 + 1 = 0\n or a3 = -1/6 \n\n12 a4 + a2 - (1/6) = 0\n or 12 a4 = (1/6) - (-1/2) = 2/3\n or a4 = 1/18\n\nSo f(x) = 1 + (-1/2) x^2 + (-1/6) x^3 + (1/18) x^4 + ...\n\nFor g it is similar except that you start with:\n\n(2 b2 x + 6 b3 x^2 + 12 b4 x^3 + ...) +\n (x - (1/3!)x^3 + (1/5!)x^5 - ...) (x + b2 x^2 + b3 x^3 + ...)\n = 0
207713
Ordinary differential equations are equations to be solved in which the unknown element is a function, rather than a number, and in which the known information relates that function to its derivatives. Few such equations admit an explicit answer, but there is a wealth of qualitative information describing the solutions and their dependence on the defining equation. \n\nThe solutions to many classic differential equations, particularly linear second-order differential equations, cannot be expressed in terms of the elementary functions. This includes Bessel functions, Whittaker functions, Airy functions, and so on.\nBuilding on tradition going back to the ancient Greek philosophers, Leibniz and Newton invented calculus in the late 17th century and thereby laid the foundation for the revolutionary development of science and technology that is shaping the world today. Calculus is a method for modeling physical systems mathematically by relating the state of a system to its neighboring states in space-time using differential and integral equations. Because calculus is inherently computational, this revolution began to accelerate tremendously in the 1940s when the electronic computer was created. Today, we are seeing what is essentially a ``marriage'' of calculus and computation in the creation of the field of computational mathematical modeling. \n\nActually, Leibniz himself sought to realize a unification of calculus and computation, but failed because the mechanical calculator he invented was not sufficiently powerful. The next serious effort was made in the 1830s by Babbage, who designed a steam powered mechanical computer he called the Analytical Engine. Again, technical difficulties and low speed stopped his ambitious plans. \n\nThe possibility of realizing Leibniz' and Babbage's visions of a universal computing machine came with the invention of the electronic valve in the 1930s, which enabled the construction of high speed digital computers. The development took a leap during the World War II spurred by the computing demands of the military. Until this time, large scale computations were performed by rooms of people using mechanical adding machines. The war provided an immediate pressure to speed up the process of scientific development by using mathematical modeling to hone a physical problem down to a manageable level, and mathematicians and physicists became interested in the invention of an electronic computing device. The logical design of programmable electronic computers was developed by the mathematician von Neumann, among others. By the late forties, von Neumann was using the first ENIAC (Electronic Numerical Integrator And Calculator) computer to address questions in fluid dynamics and aerodynamics. \n\nThe subsequent development of computer power that has resulted in desktop computers of far greater power than the ENIAC, has been paralleled by the rapid introduction of computational mathematical modeling into all areas of science and engineering. Questions routinely addressed computationally using a computer include: What is the weather going to do in three days? Will this airplane fly? Can this bridge carry a load of ten trucks? What happens during a car collision? How do we direct a rocket to pass close by Saturn? How can we create an image of the interior of the human body using very weak X-rays? What is the shape of a tennis racket that has the largest "sweet spot"? What is a design of a bicycle frame that combines low weight with rigidity? How can we create a sharp picture from a blurred picture? What will the deficit be in Sweden in the year 2000? How much would the mean temperature of the earth increase if the amount of carbon dioxide in the atmosphere increased by 20 percent? \n\nThe physical situations behind these kinds of questions are modeled by expressing the laws of mechanics and physics (or economics) in terms of equations that relate derivatives and integrals. Common varia
120626
There's a derivatives tutorial on the following page:\n\nhttp://www.karlscalculus.org/calculus.html#s4