Can someone explain this to me please?

AdrianZ

I'm asking this integral equation (I'm not sure if it's an integral equation or not by it's a problem in my ODE book and because it has an integral in it I called it that way). anyways, this is the problem:

[tex]y=\int^{x}_{1}ty(t)dt[/tex]

I differentiated y with respect to x and I turned that equation into this ODE: [tex]y'=xy[/tex]
Solving this ODE yields [tex]y=Ce^{x^2/2}[/tex]

But from the definition of y, it is clear that y(1)=0 while my solution suggests that y=e^1/2.

Then I substituted y(t)=Ce^x2 in the original equation and I obtained:
[tex]y=\int^{x}_{1}tCe^{t^2/2}dt → y=C(e^{t^2/2})|^{x}_{1}→y=C(e^{x^2/}-e^{1/2})[/tex]
And in this case y(1) is indeed equal to 0.

Would someone explain why the y that is obtained from the ODE solution tells me that y(1)≠0? What's wrong in my solution?

AdrianZ

Any ideas on this matter will be appreciated. Are my questions really that hard that they usually get no responses back on physics forum or there's a conspiracy against me? lol.

omkar13

The condition y(1)=0 is given in order to find constant.THE PURPOSE OF GIVING BOUNDARY CONDITIONS IS TO OBTAIN CONSTANTS AFTER INTEGRATION.

yus310

"In order to solve this you need to know what value of t... x and 1 correspond to and then you can go from there. If you keep it in terms of t, you get the same function, but t^2, x^2.

Knowing that x=t, at y=?, and t=0, y=?, you can transform accordingly .

This an ODe, you you don't have what y=, when x=t, t=0, etc.

YS

AdrianZ

I guess I've already found an explanation to this.
C must be zero. probably that's the only answer this integral equation can have. other answers would lead to contradiction.