Change of variables, often a differential equation with variable coefficients, y

Question

Change of variables, often a differential equation with variable coefficients, y" + p(t)y' + q(t)y = 0 can be put in a more suitable form for finding a solution by making a change of the independent and/or dependent variables. Let us determine conditions on p and q that enable this equation to be transformed into an equation with constant coefficients by a change of the independent variable. Let X = u(t)be the new independent variable. It is easy to show that dy/ dt = dx/dy dy/dx, d2y/dx2 = (dx/dt)2 d2y/dx2 + d2x/dt2 dy/dx (dx/dt)2 d2y/dx2 + (d2x/dt2 + p(t) dx/dt) dy/dx + q(t)y = 0 In order for to have constant coefficients of d2y / dx2 and y must be proportional. If q(t) > 0, then we can choose the constant of proportionality to be 1 ; hence With X chosen this way, the coefficient of dy/dx in is also a constant, provided that the function H = q(t) + 2p(t) q(t) / 2[q(t)]3/2 is a constant, thus can be transformed into an equation with coefficients by a change of the independent variable, provided that the function H is a constant. Try to transform the given equation into one with constant coefficients by this method. Is it possible y" + 4ty' + t2y = 0

Explanation / Answer

v'' + 4tv' + t^2v = 0 p(t) = 4t q(t) = t^2 now, x = integral(sqrt(q(t))dt = integral(tdt) = t^2/2 => x = t^2/2 and t = sqrt(2x) now, dx/dt = t and, d^2x/dt^2 = 1 now, our equation transforms into : (dx/dt)^2V'' + (d^2x/dt^2 + p(t)dx/dt)V' + q(t)V = 0 t^2V'' + (1 + 4t^2)V' + t^2V = 0 V'' + ((1+4t^2)/t^2)V' + V = 0 no, it was not possible to convert all coefficients into constants since coefficient of V' is not a constant

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