Given that cos x and e x are solutions to the non-linear equation (y \'\' ) 2 -

Question

Given that cos x and ex are solutions to the non-linear equation (y'')2 - y2 = 0. Verify that sin x and e-x are also solutions. Without attempting to solve the differential equations, discuss how these explicit solutions can be found by using knowledge about linear equations. Without attempting to verify, discuss why the linear combinations y = c1ex+c2e-x+c3cos(x)+c4sin(x) and y = c2e-x +c4sin(x) are not, in general, solutions, but the two special linear combinations y = c1ex + c2e-x and y = c3cos(x) + c4sin(x) must satisfy the differential equation.

Explanation / Answer

A linear ODE in t and y has the form

Sum[j=0,n; f(t;n) (d^j y) / (dt)^j] = F(t)

The first is non-linear because the y^2 doesn't fit the form.

The second is linear because it fits the form.

Essentially, the form is linear when does not contain products of the derivatives of y. [y^2 = product of the zeroth order derivate by itself.]

As with a Taylor series, you linearize about a particular value of x, call it p. This is crucial! You linearize ***about a particular point***. Typically you linearize about a fixed point (a.k.a. steady state or equilibrium point) of the system, i.e. where f(p) = 0 for some value(s) x=p.

i.e. f(p+x) = f(p) + f'(p)*x + higher order terms

so you get dy/dx = f(p) + f'(p) * x as your linearized system. In classes and textbooks you're often given that the equilibrium has been moved to p=0 for extra simplicity (it can always be translated to be at the origin without loss of generality) so that f(p)=0, and so you might see dy/dx = f'(p) * x written instead.

To be a little more general, if x is in R^n (i.e. is a vector), then you take the Jacobian of f: R^n -> R^n instead of the usual 1D derivative.

duce to a first-order, nonlinear, ordinary equation, but I don't think there is a closed-form solution that is expressible in terms of elementary functions.

let u(y) = dy/dx

then

d^2y/dx^2 = du/dx = du/dy * dy/dx = u*du/dy

This transforms the equation to:

u*du/dy = y/(1+y)

This is now a separable equation:

u du = (y*dy)/(1+y)

u du = (1 - 1/(y+1)) dy

(u^2)/2 = y - ln(y+1) + c

where c is the constant of integration.

u^2 = (2y - 2ln(y+1) + C)

u = +sqrt(2y - 2ln(y+1) + C) and u = -sqrt(2y - 2ln(y+1) + C)

dy/dx =

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