Question

ODE problem

Implicit equation.2 Consider the ODE (a) Solve the equation by first obtaining explicit equation(s) for y. Here y : R → R is a scalar function. Your answer should be a family of solution parameterized by a single (b) Show that the function y(x) 0 also solves the equation even though it does not belong (c) Sketch a few representative functions of the family found in part (a) together with the parameter (an integration constant). to the family found in (a). Solution such as this are called singular solutions. singular solution of part (b). What is the geometrical relation between these solutions? (d) Is the solution to the ODE (11) with a given initial condition y(0) yo unique?

Answer 1

Given ODE is (y'(x)^2 -4y(x) = 0 ----(*) (a) y(x) = (x + c)^2: c elementof R, then y'(x) = 2(x + c) (y'(x))^2 -4y(x) = (2(x + c))^2 -4(x + c)^2 = 0 Therefore y(x) = (x + c)^2: c elementof R is a general solution of (*). (b) y(x) = 0, then y'(x) = 0 (y'(x))^2 -4y(x) = (0)^2 -4(0) = 0 Therefore y(x) = 0 is a solution of (*) which does ot belongs to the family y(x) = (x = c)^2: c elementof R Here y = 0 is envelope of the family of curves y = (x + C)^2, c elementof R. That is y = 0 is tangent to a family of curves y = (x + c)^2, c elementof R. (d) y(0) = y_0, y_0 > 0 implies y_0 = c^2 implies c = plusminus squareroot y_0 Therefore y(x) = (x plusminus squareroot y_0)^2 Hence the solution of the ODE (*) with