First order linear ode, initial condition problem

1. Apr 26, 2012

linda300

hey,

i'm having trouble with this question,

x y' - y = x2cosx

the solution is

y= xc + xsinx

and we are asked to solve the equation in the following two cases,

1, y(0)=0

and 2, y(0) = 1

but applying these conditions to the general solution gives no information,

in both cases its 0 = 0, so doing this you cant find the constant.

is there another way to solve the equation for these conditions?

even mathematica can't solve it, but it was a question in a test and i was marked wrong for saying neither cases can be solved (as in the constant cant be found)

thanks

2. Apr 26, 2012

HallsofIvy

Staff Emeritus
Has it occured to you that the whole point of this exercise is that there is NO unique solution?

I suspect that you either have just had or will soon have the "Fundamental existance and uniqueness theorem" for first order equations:

If f(x,y) is continuous and "Lipschitz in y" (a condition that is partway between "continuous" and "differentiable"- "differentiable" is sufficient and many texts just use "differentiable in y") in some region around the point $(x_0, y_0)$ then there exist a unique function, y(x), satisfying dy/dx= f(x,y) and $y(x_0)= y_0$.

In this example, solving for y' gives $y'= (x^2 cos(x)- y)/x$ so that $f(x,y)= (x^2 cos(x)- y)/x$ is not continuous in any neighborhood containing x= 0. Thus, the "existance and uniqueness theorem" does not apply. Your first example is one in which a solution exists but is not unique. Your second is one in which a solution does not exist.