Why do different methods give different solutions for this initial value problem?

[malee006]

Hey ! i am having a problem in differential initial value problem, can it happen that solving an initial value problem with two different methods (e.g. method of undetermined coefficient and using Laplace) give us two different answer!

the equation is : y"-4'=(5e^-2t)+1 y(0)=0 y'(0)=10


and if possible can u do it by method of undetermined coefficient bcoz i think i am doing a mistake in tht method becoz of that '1' on right hand side and no comparing coefficient on the other side!

 

[rock.freak667]

$$\frac{d^2y}{dt^2}-4=5e^{-2t}+1$$

Aux.Eq'n=$$r^2-4=0 \Rightarrow (r+2)(r-2)=0 \Rightarrow r= \pm2$$

$$y_{CF}=Ae^{2t}+Be^{-2t}$$

I think you got this far right?

Well since -2 is one root of the aux. eq'n. The PI for the exponential should be xe^-2t

 

[Alex6200]

Yeah, I agree. Try laplace transform method and for the second method find the roots of the characteristic equation.

 

[HallsofIvy]

I personally think the "Laplace Transform method" is overkill.

Malee006, I presume the equation is y"- 4y'= 5e^2t+ 1. Then, as Rockfreak667 said, the general solution to the homogeneous equation is Ae<sup>2t[/sub]+ Be-2t[/itex].

Since e2t is already a solution, you need to look for a specific solution to the entire equation of the form Cte2t+ B. The "B" is to get the "1" on the right hand side of the original equation.</sup>