sketch an approximate solution of the initial value problem

sketch an approximate solution of the initial value problem dy/dx=y cos(y), y(0)=1

Solution

A differential equation is an equation that contains an unknown function and one or more of its derivatives. Here are some examples: y 0 = 1, y0 = x, y0 = xy y 00 + 2y 0 + y = 0 d 3 y dx3 + x d 2 y dx2 + dy dx 2y = e x The order of a differential equation is the order of the highest derivative that occurs in the equation. A function f is called a solution of a differential equation if the equation is satisfied when y = f(x) and its derivatives are substituted into the equation. For instance, we know that the general solution of the differential equation y 0 = x is given by y = 1 2 x 2 + C, where C is an arbitrary constant. Separable Equations A separable equation is a first-order differential equation that can be written in the form dy dx = g(x)f(y) The name separable comes from the fact that the expression on the right side can be “separated” into a function of x and a function of y. Equivalently, if f(y) 6= 0, we could write dy dx = g(x) h(y) (1) where h(y) = 1/f(y). To solve this equation we rewrite it in the differential form h(y)dy = g(x)dx so that all y’s are on one side of the equation and all x’s are on the other side. Then we integrate both sides of the equation: h(y)dy = g(x)dx (2) Equation 2 defines y implicitly as a function of x. In some cases we may be able to solve for y in terms of x. We use the Chain Rule to justify this procedure: If h and g satisfy (2), then d dx ( h(y)dy) = d dx ( g(x)dx) so d dy ( h(y)dy) dy dx = g(x) and h(y) dy dx = g(x) 1 Section 7.6 Differential Equations: Direction Fields EXAMPLE: (a) Solve the differential equation dy dx = x 2 y 2 . (b) Find the solution of this equation that satisfies the initial condition y(0) = 2. Solution: (a) We write the equation in terms of differentials and integrate both sides: y 2 dy = x 2 dx y 2 dy = x 2 dx 1 3 y 3 = 1 3 x 3 + C hence y = 3 x 3 + 3C which can be rewritten as y = 3 x 3 + K where K = 3C. (b) If we put x = 0 in the general solution in part (a), we get y(0) = 3 K. To satisfy the initial condition y(0) = 2, we must have 3 K = 2 and so K = 8. Thus the solution of the initial-value problem is y = 3 x 3 + 8 EXAMPLE: Solve the differential equation dy dx = 6x 2 2y + cos y . Solution: Writing the equation in differential form and integrating both sides, we have (2y + cos y)dy = 6x 2 dx (2y + cos y)dy = 6x 2 dx y 2 + sin y = 2x 3 + C where C is a constant. The last equation gives the general solution implicitly