# 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.