Question

3 Consider the ordinary differential equation: ty +3tyy 0.

Answer 1

The given differential equation t^2y'' + 3ty' + y = 0 i.e., t^2 d^2y/dt^2 + 3t dy/dt +y = 0 if we take dy/dt = Dy and d^2y/dt^2 = D^2y, then 1 becomes t^2D^2y + 3t Dy + y = 0 (t^2D^2 + 3tD + 1)y = 0 let t = e^x so that , dt = e^x dx dx/dt = 1/t and t dy/dt = e^x dy/dx dx/dt = e^x dy/dt dx/dt = e^dy/dx 1/e^x = dy/dx = D_1y t^2d^2y/dt^2 = e^2x [d/dt [1/t dydx]] = e^2x [-1/t^2 dy/dx + 1/t d^2y/dx^2 dx/dt] = d^2y/dx^2 - dy/dx = D^2_1 y - D_1 y (says) \r\n therefore lt_t rightarrow infinity y(t) = lt_x rightarrow 0^+ ln(1/x) /(1/x) = lt_x rightarrow 0 1/(1/x) (-1/x^2) /(-1/x^2) = lt_x rightarrow 0 x = 0 lt_t rightarrow infinity y(t) = 0