5 Let f be a realvalued continuous function defined on the r

5. Let f be a real-valued continuous function defined on the rectangle Let o be a real-valued function defined on an interval I containing zo. (a) Define carefully what it would mean to say that is a solution on 1 of the initial value problem (b) Define carefully what it would mean to say that is a solution on l of the integral equation v(x -z,y) dt. #0 (c) Show that is a solution of the initial value problem (*) on 1 if and only if is a solution of the integral equation (**) on I. (Hint: In proving the statement one way it is useful to use the rule that dz Ja a dx if F and aF/ax are continuous. In proving the statement the other way, let F(z)-f(x, (z) ), and solve by the variation of constants method.)

Solution

Given that is a real valued function defined on the interval I containing x₀

y\'\' = f(x,y), y(x₀) = y₀, y\'(x₀) = y₁  

Now y\'\' = f(x,y)

Integrating both sides, we

y\' = f(x.y) dx + c₁

Again Integrating we get

y = (f(x.y) dx)dx +  c₁ x + c₂

The constants c₁, c₂ can be calculated by putting the values at x = x₀ and y = y₀

So, defined on the interval I containing x₀ will be a particular solution on y.

(c) After obtaining the values of C₁ and C₂ we get the given y

i.e. y = y₀ + (x - x₀) y1 + (x - t ) f(t, y) dt