Initial value problem Definition and 172 Threads

x=c_1\cos{t}+c_2\sin{t} is a two-parameter family of solutions of the DE x''+x=0 Find a solution of the IVP consisting of this differential equation and the following initial conditions: x(\frac{\pi}{6})=\frac{1}{2} and x'\frac{\pi}{6}=0 So x'=c_2\cos{t}-c_1\sin{t} x''=-c_2\sin{t}-c_1\cos{t}...

Homework Statement I'm stuck trying to find out the inverse Laplace of F(s) to get y(t) (the solution for the differential equation): Y(s) = 1 / [ (s-1)^2 + 1 ]^2 The Attempt at a Solution I tried using a translation theorem and then apply the sine formula, but the denominator...

Homework Statement Solve the initial value problem: dx/dt = x(2-x) x(0) = 1 Homework Equations Problem statement. The Attempt at a Solution Based on the format, I attempted to solve the problem as a separable differential equation: ∫dx/(x[2-x]) = ∫dt Evaluating to...

Solve the initial value problem y'=2t(1+y), y(0)=0 by the method of successive approximations. I don't know how to do this problem but I think there's integral involved in it. Please help me. Thanks.

Homework problem for nonlinear dynamics. Let us write xλ(t) for the solution of the initial value problem [SIZE="4"]\dot{x} = f(x) & x(0) = λ where f is continuously differentiable on the whole line and f(0) = 0. a) Find the differential equation for [SIZE="4"]\frac{∂x_{λ}}{∂λ}(t)...

Here is the question: Here is a link to the question: Calculus question on differential equations? - Yahoo! Answers I have posted a link there to this topic so the OP may find my response.

Homework Statement 1.)I want to write a function in MATLAB that contains the 2nd order function: 20*d^{2}x;(dt^{2})+5*dx/dt + 20*x=0 (dampened spring) -The function should have 2 inputs (time,[initial values]) initial values should be a vector of 2 values -The function should...

Homework Statement Find a solution to the initial value problem that is continuous on the interval where Homework Equations I know the equations, but don't want to type them out.The Attempt at a Solution I got the first part of this question. The part where g(t) = sin(t) I can not figure...

Homework Statement The unique solution to the initial value problem is http://webwork.usi.edu/webwork2_files/tmp/equations/ed/12ad7dca5df62ed3b18f5fbf8c6e871.png Determine the constant and the function Homework Equations Not sure for the second part. The Attempt at a...

Homework Statement y''+4y'+6y y(0) = 2; y'(0) = 4 Homework Equations \alpha ± β = e^{x\alpha}(cosβx + sinβx) The Attempt at a Solution Auxilary equation is r^2+4r+6, which solves for -2 ± i I get the general solution: e^{-2x}(c1cosx + c2sinx) y' = -2e^{-2x}(c1cosx +...

I've had to take diff eqtns now and I'm trying to get my head around Laplace again.. it's been a while. I can't seem to transition to the simplest step of partial fractions, my denominators are tough to figure out. If someone could point me to the next step that'd be great! Thanks a lot guys...

Hello. I have gotten as far as to use the Laplace equation with these formulas, but I am having difficulty getting y and x to relate to each other. If requested, I can post my work, but I am sure it is fraught with mistakes. Help is very much appreciated! x' + 2y' - x - 2y = e^t x' - y' + x...

Hey everyone, I'm a long-time visitor, it's my first time posting though. I have a homework problem that is causing me considerable consternation: (y^3)*(dy/dx)=(8y^4+14)*cos(x); y(0)=C Oh, and we're supposed to solve the initial-value problem, and then solve for the particular...

Homework Statement For the space of continuous functions C[0,T] suppose we have the metric ρ(x,y) =sup _{t\in [0,T]}e^{-Lt}\left|x(t)-y(t)\right| for T>0, L≥0. Consider the IVP problem given by x'(t) = f(t,x(t)) for t >0, x(0) = x_{0} Where f: ℝ×ℝ→ℝ is continuous and globally Lipschitz...

Homework Statement y'' +4y = 2 delta(t - pi/4) where y(0)=0 and y'(0)=0 Homework Equations Laplace transform Inverse Laplace transform The Attempt at a Solution after applying laplace tranform Y(s)=2e^((-pi/4)*s) / s^(2)+4 as the final answer i have y(t) =...

Homework Statement y''-4y'+4y=0 , y(1)=1 and y'(1)=1 The Attempt at a Solution Auxiliary equation: r2-4r+4=0 I tried factoring 2 different ways: (r-2)2=0 r=2,r=2 y1=e2t y2=y1 y(t)=c1e2t+c2e2t y(1)=c1e2+c2e2=1 ---eq(1)y'(t)=2c1e2t+2c2e2t ...c2=1/(2e2)-c1 ---eq(2) sub eq(2) into eq(1)...

Given: Solve the initial value problem 2(√x)y'+y+4(√x) ; y(1)=2 I am having trouble separating the x's and y's in order to integrate. I keep coming up with: dy/dx +y/(2(√x))=2... What do I keep missing here? I am pretty sure you leave the y(1)=2 alone until you are finished with...

Homework Statement Solve the following Initial Value problem for x(t) and give the value of x(1) Homework Equations (dx/dt)-xt=-t , x(0)=2 The Attempt at a Solution (dx/dt)-xt = -t (dx/dt) = xt-t (dx/dt) = t(x-1) (1/(x-1)) (dx/dt) = t (1/(x-1)) dx = t dt Then I integrate...

how can we solve this ODE? http://img818.imageshack.us/img818/3966/59962234.png

Thanks for clicking! So, I've got a problem here that I'm stuck on. I need to find the general solution to y' = (y3 + 6y2 + 9y)/9 I found this to be ln|y| + (3/(y+3)) - ln|y+3| = x + c but I would appreciate it if you would check my work. Anywho, once I have the general solution I...

y' = x x' = -5y-4x y(0) = 1 x(0) = 0 after finding the general solution as shown here http://www.wolframalpha.com/input/?i=y%27+%3D+x%2C+x%27+%3D+-5y-4x how do you go about applying the initial values and finding the complete solution?

L\frac{dI}{dt}+RI=E I(0) = I_{0} Where E is a constant. I know I need to separate the equation and integrate but I am not quite sure how given all the variables running around... I don't see how the condition of I(0) = I_{0} helps in any way.

The equation is y'' + 4y' + 4y = (3 + x)e-x and initial conditions y(0) = 2, y'(0)=5so from the associated homogenous equation I think the fundamental set of solutions is {e^-2x, xe^-2x} and so yc would be Yc = c1e-2x + c2xe-2x but now I don't know how to get Yp, particular solution or what...

Homework Statement Determine the solution of the IVP y' + 4ty = 4t, y(0) = 6 Homework Equations The Attempt at a Solution p(t) = 4t g(t) = 4t μ(t) = e^{\int4tdt} = e^{\int p(t)} = e^{\int4tdt} = e^{2t^{2}} is this all I need? because i did \frac{d}{dt}(y * μ(t)) = p(t)...

dy/dt=t^(2)y^(3) , y(0)=-1 I need help solving this I put the integral (dy/y^3)= integral (t^2)dt but idk what to do after that or if that's even right

Homework Statement R(dQ/dt) + (1/C)Q = E_0 e^-t ...Q(0) = 0 and E_0 = a constant Homework Equations The Attempt at a Solution first i rearranged to give: Q' + (1/CR)Q = (E_0e^-t)/R next i multiplied all by integrating factor of: u(t) = e^integ:(1/CR) = e^(t/CR)...

Homework Statement Solve the I.V.P. x2(dy/dx) = (4x2-x-2)/((x+1)(y+1)) , y(1)=1 Homework Equations The Attempt at a Solution So far, I got to this: y2/2 + y = log(x) + 2/x + 3log(x+1) + C I used the initial conditions to solve for C and got: C = -1/2 - 3log(2) Substituting C...

OK, so clearly I am missing something, because I know this is supposed to be a simple problem. It reads: solve the following initial value problem: dy/dt=-y+5 y(0)=y_naught my process is as follows: dy/(5-y)=dt integrate ln(5-y)=t+C exponential both sides 5-y=(e^t)(e^c)...

Homework Statement I understand how to do initial value problems but I'm slightly stuck when the initial values are y(0) = y'(0)=0 The question is Solve: y''+3y''+2y=f(t), y(0)=y'(0)=0 where f(t) is a square wave. Homework Equations \Im{y'} =s\Im{y}-y(0)...

Homework Statement We know that y = Aex is the solution to the initial value problem dy/dx = y; y(0) = A. This can be shown by solving the equation directly. The goal of this problem is to reach the same conclusion using power series. Method: Let y be a solution to the initial value...

Homework Statement 3y'' -y' + (x+1)y = 1 y(0) = y'(0) = 0 Homework Equations Not sure, that's the issue The Attempt at a Solution I can't quite get this one using the methods I'm familiar with, and I can't guess a particular solution to neither the equation nor the...

Question: Find y as a function of x: x^2 y'' + 8 x y' - 18 y = x^8 y(1)=3, y'(1)=2 Attempted solution: I found the general equation to be Ax^(-9)+Bx^2+Cx^8. However when I try to solve the initial value problem for this equation I have 3 unknowns.

I'm given the following DE and initial conditions: y''=2yy' y(0)=0, y'(0)=1 I started by doing a reduction of order like so: w=y', w'=y'', \int w=y=\frac{w^{2}}{2}+c which then gave me this: w'=2w(\frac{w^{2}}{2}+c) w'=w^{3}+2wc Now I'm stuck on where to go from here. I can't use any of the...

Homework Statement Find the solution to the initial value problem dy/dx - y = e^3x y(0) = 3 Homework Equations e^∫p(x) The Attempt at a Solution Do I treat p(x) = -1? I(x) = e^∫-1 = e^-x e^-x(dy/dx) - ye^-x = e^3x . e^-x e^-x(dy/dx) - e^-x . y = e^2x e^-x . y = ∫e^2x y = (2e^2x...

Homework Statement Solve the Initial value problem using Laplace transform \ddot{y} +2y = 0, y0 = C1, \dot{y} = C2 Homework Equations [s2 - sy(0) - y'(0)] + a[sY - y(0)] + bY The Attempt at a Solution s2Y - sy(0) - y'(0) + 2y = 0 s2Y + 2Y = sy(0) + y'(0) (s2 + 2)Y = s(C1) + (C2)...

So the question is y" - y' - 6y = e^-x + 12x, y(0)=1,y'(0)=-2 First I found the general solution which came out to be, Ae^3x + Be^-2x I then Substituted y=ae^-x + bx + c y'=-ae^-x + b y"=ae^-x Then I just compared the coefficients to get a=-1/4, B=-2 and C=-1/6 So I am getting y =...

a 3rd order IVP I am havin trouble with: y''' -3y'' +2y' = t + e^t y(0)=1, y'(0)= -.25 y''(0)= -1.5 I am using At^2 and B*e^t *t as my Y1 and Y2. Is this correct?

Hey, we have to solve the following problem for our ODE class. Homework Statement Find the solution of the initial value problem dx/dt = (x^2 + t*x - t^2)/t^2 with t≠0 , x(t_0) = x_0 Describe the (maximal) domain of definition of the solution. The Attempt at a Solution Well...

1. y'' -5y' + 6y = 0, y(0) = 1, y'(0) =2 3. [s^2 F(s) - s f(0) - f'(0)] -5 [F(s) - f(0)] + 6[F(s)] = 0 (s^2 +1)F(s) - (s -5)f(0) - f'(0) = 0 (s^2 + 1)F(s) - (s-5)(1) - 2 = 0 (s^2 + 1)F(s) = s -3 F(s) = (s-3)/(s^2 + 1) Here's where I'm stuck. I can't factor the denominator to...

I am having trouble with the below problem: y'-(3/2)y= 3t+ 2e^t, y(0)= y0. fine value of y0 that separate solutions that grow positively and negatively as t=> infinity. I found p(t) to be -3/2, u(t) to be e^-3t/2 => e^-3t/2*y' - 3y/2( e^-3t/2)= e^-3t/2(3t+ 2e^t) => -2 -4e^t + ce^...

I need help with an initial value problem, ty' + (t+1)y= t; y (LN 2)= 1 I divided t and have u(t) as exp Integral of t+1/1 => e^t +t Multiplied this to the original equation to get (e^t +t)y' + ((t+ 1)/t) *y *(e^t +t) = (e^t +t) How can I integrate the above? Are my steps so far...

Homework Statement Proof that there exist more than one solution to following equation \frac{dx}{dt} = \sqrt[3]{x^{2}} , x(0) = 0Homework Equations The Attempt at a Solution Well, I need a confirmation to my attempt of solution. The one is quite forward: \Rightarrow x=(1/3(t+c))^{3} Pluging...

Homework Statement Find the general solution for the following systems of equations, a solution to the initial value problem and plot the phase portrait. --> this is in matrix formx' = 1 2 0 3 all multiplied by x. also, x(0) = 2 -1 Homework Equations Determinant, etc.The Attempt at...

Is this problem possible? Solve the initial value problem x''(t) + 6x'(t) + 9x(t) = f(t); x(0) = N, x'(0) = M I get to X(s)=(F(t)+Ns+6N+M)/(s^2+15) and don't know where to go from here. Any help would be appreciated.

Hey, I need some guidance on an IVP. In general, how do you proceed on these types of problems when you have only the initial values but no initial equation? For example, I have x1(0)=1 and x2(0)=0 but that is it. I understand, for example, how to do IVP's in the context of separating...

Homework Statement solve the initial value problem for u du/dt= (2t + sec^2(t))/2u also, u(0)=4 Homework Equations antiderivative of sec^2(t) is tan(t) + C The Attempt at a Solution So, the first thing i did was move the "u" with the "u" and "t" with the "t". so the equation looks like...

I am working on the following initial value problem: y'' + 4y = 4, y(0) = 1, y'(0) = 1 The method they show is: (1): y = A cos x + B sin x A = 1, (2): y = 1 + A cos x + B sin x y(0) = 1 + A = 0 y'(0) = B = 1 Final answer is y(x) = 1 + sin(x) The problem I don't see is the first...

Homework Statement dx/ dt = x + y dy/ dt = x + y + et x(0) = 0 y(0) = 1.Homework Equations The Attempt at a Solutionx'' = x' + y' = x' + x + y + et x'' - 2x' = 0 y = x' - x

Homework Statement y' = (2x) / (y+(x^2)y) y(0) = -2 The Attempt at a Solution I tried doing this by finding the Integrating factor and I got that to be u = -1-x^2 by using the (My - Nx) / N formula. Using this did not work out for me and I'm not seeing the other approach...

Homework Statement d2y/dx2 = 2-6x Given: y(0)=-3 and y'(0)=4 Homework Equations None that I know of. The Attempt at a Solution I know that for a single order derivative you would just find the integral, set y=1 and x=0. But I'm confused because here we're given two...