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I think you can finish it now. In addition to Julián's answer, note that the LHS is precisely the derivative of x2y(x) x 2 y (x), which readily leads to the change of variables that he pointed out. Thus, you have: dz dx =(z x2)2, d z d x = (z x 2) 2, which can now easily solved. Cheers! Note that. x2y′(x) + 2xy(x) = [x2y(x)]′ x 2 y ′ (x ...
A steady state solution is a solution for a differential equation where the value of the solution function either approaches zero or is bounded as t approaches infinity. It sort of feels like a convergent series, that either converges to a value (like f(x) approaching zero as t approaches infinity) or having a radius of convergence (like f(x ...
The theorem of existence and uniqueness is: Let $ y'+p(x)y=g(x) $, $ y(x_{0})=y_{0} $ be a first order linear differential equation such that $ p(x) $ and $ g(x) $ are both continuous for $ a<x<b $. Then there is a unique solution that satisfies it. When a differential equation has no solution that satisfies $ y(x_{0})=y_{0} $, what does this ...
Depending on why you need a first-order differential equation, this might be the answer you're searching for, and it might not. It's what we did in introductory numerical solving of differential equations, and it works if what you want to do is to apply the Euler method or Runge-Kutta, or something of the sort.
A Linear First Order Differential Equation is of the form: dy dx + P(x)y = Q(x) Where P(x) and Q(x) are only functions of the independent variable (in this case x) I do not want to go into the details of an integrating factor so, the general solution to such an equation is ye∫, where C is an arbitrary constant and the e ∫ P (x) dx is known ...
Express the 2nd order ODE $$\\begin{align}\\mathrm d_t^2 u:=\\frac{\\mathrm d^2 u}{\\mathrm dt^2}&=\\sin(u)+\\cos(\\omega t)\\qquad \\omega \\in \\mathbb Z /\\{0 ...
$\begingroup$ This is late question, but what about singular solutions ? first order differential equation should has 1 solution, but some have more singular solutions. ? $\endgroup$ – Mohamed Mostafa
You can apply this same method to your other differential equation $\frac{dy}{dx}-\frac{y}{x}=1$ by letting $1$ equal $0$ to find a solution to your homogeneous equation. Share Cite
The singular solution is basically like an envelope of all the solutions of the differential equation. To find the general solution, we do the following, let p = dy dx p = d y d x. The differential equation is. y − px −p2 = 0 y − p x − p 2 = 0. Differentiating just like you did,
If the constant gets cancelled throughout and we obtain the same equation again then that particular differential equation is homogeneous and the the power of constant which remains after cutting it to lowest degree is the degree of homogeneity of that equation.