2nd order Definition and 464 Threads

RLC second order linear network question: So, we are given this equation which is the same for Vc(t) and iL(t) expressed as x(t): 2nd deriv of x(t) + R/L(1st deriv of x(t)) + 1/(LC)(x(t)) = 0; And in one of the problems it asks to find both equation for the Vc(t) and iL(t) for t < 0, and...

I am trying to match a result in one of my textbooks. To assist with one of their arguments they are approximating a 2nd order PDE by using a difference quotient and they show the approximation as follows: (d^2u[x,t])/(dx^2) =~ (1/h^2)(u[x+h,t]-2u[x,t]+u[x-h,t]) When I actually use...

I have this DE and I do not know how to solve it. This is a problem in my review questions that my professor handed out, but we have never done a problem like this in class, so I do not know where to start. here is the problem: x'' = 4x - 5y y'' = 2x - 3 I need to solve it using...

I'm stuck as to where to start with this question: The position function x(t) in a certain nonlinear system is described by the second order ODE: < equation.gif > Transform this ODE into a pair of first order ODEs for x1=x and x2=dx/dt. (Note that x2 represents the velocity in this...

this is the problem: xy'' -x(y')^2 = y' my book says that i need to substitute u=y' and du/dx=y''... so i get: x(du/dx)-xu^2 = u so next the book says i need to separate the x's/dx' to one side and u's/du's to the other. however, i cannot do it am i using the correct technique...

problem: xy'' -x(y')^2 = y' what i have so far: u=y' and du/dx=y'' du/dx - u^2 = (1/x)u int[(1/u)-u]du = int[1/x]dx ln u - (1/2)u^2 = ln x +c ok, now is what I've done so far correct? what do i do next? ps: i'd like to say hi to everyon :) I am new here

Hi Everybody, Does anybody know how to solve, analytically or numerically, the following differential equation : \frac{d^2\Phi}{dx^2}-a.Sinh(\frac{\Phi}{U_{th}})=-b.Exp(-(\frac{x-x_{m}}{\sigma})^2}) The unknown function is \Phi. a and b are some strictly positive constants. q\Phi is...

A theorem in my textbook is confusing me: For the functions p(t) \ \ \text{and} \ \ q(t) continuous on an open inteval I defined by \alpha < t < \beta : We have differential equation L[y] = 0 where L = (\frac{d^2}{dt^2} + p\frac{d}{dt} + q) The theorem attempts to prove...

Help! I'm just starting this class and I have no idea what's going on. What I don't understand is, what answer are you supposed to give? My question says "Find the general solution and also the singular solution, if it exists". What the hell does that mean? Can someone tell me if this...

xy'' -(2x+1)y' + (x+1)y = (x ex)2 I know a solution - (x-1)e2x Thus, y= ((x-1)e2x u(x)) Now, i know how to do the whole reduction of order thing, but when i find y' and y'' and substitute, the u(x) term doesn't cancel out so this doesn't work (x2-x)u'' + (2x2-x+1)u' + x2u = x2 So...

After setting out in the sums and collecting the terms in x^j I'm left with a series of expressions in a_2, a_3 etc as I believe I'm supposed to. However my first expression reads 2a_{2}+2a_{1}+a_-_{1}=0 Now I'm told that y(o) = 1 and y'(o) = 0 I think this means that a_0 = 1 and...

It looks simple enough: y'' + x*y = x^2 However, I tried and I could not find a nontrivial solution to the homogeneous equation: y'' + x*y = 0 Am I right in thinking you need to solve this with series? No need to actually do it, I just need to know if it is possible otherwise...

ok im trying to solve the following equation using standard aux method: d^2y/dx^2 + 3dy/dy +2y = cos x with conditions x(0)=-3 and x'(0) = 3 my aux eqn is: ae^x + be^-2x and my yp is; a sin kx + b cos kx i differentiate this twice and substitute into the original...

hi all... i need to solve this differential equation of 2nd order...if anyone could tell me the way or method to use to do it i would appreciate it. i don't mind if the sol is given :wink: a*x^2 y''+(bx-c1)*y'-by+c2=0 where y'=dy/dx and y"=d^2y/d^2x a,b,c1,c2 are constants thanx