Question

problem 34

Equations with the Independent Variable Missing. If a second order differential equation has the form y"f(y, y), then the independent variable t does not appear explicitly, but only through the dependent variable y. If we let y', then we obtain dv/dt-f(y, v). Since the right side of this equation depends on y and v, rather than on and v, this equation is not of the form of the first order equations discussed in Chapter 2. However, if we think of y as the independent variable, then by the chain rule dv/dt- (dv/dy)(dy/dt)-v(dv/dy). Hence the original differential equation can be written as u(du/dy) /(y, u). Provided that this first order equation can be solved, we obtain v as a function of y. A relation between y and results from solving dy/dt-(y). Again, there are two arbitrary constants in the final result. In each of Problems 34 through 39 use this method to solve the given differential equation. 34. yy" (y) o

Answer 1

"yy"" + (y')^2 = 0 Let v = y' then from (1) we have yv' + v^2 = 0 Now, if we think y as independent variable by chain rule dv/dt = du/dy middot dy/dt = v dv/dy (as v = y') dv/dt = v' = v dv/dy then from (2) we have y. u dv/dy + v^2 = 0 y + v + v dy = 0 d(vy) = 0 integrating, we have vy = c_1, c_1 being arbitrary constant. V = c_1/y dy/dt = c_1/y (as v = y' = dy/dt) y dy = c_1 dt integrating, we have y^2/2 = c_1 t + c_2, c_2 is arbitrary constant \r\n y^2 = 2c_1 t + 2c_2 = d_1 t + d_2 (d_1 = 2c_1, d_2 - 2c_2) thus the general solution of (1) is y^2 = d_1 t + d_2, d_1 and d_2 are arbitrary constants. y"" + y = 0 Let v = y', then from (3) we "