Inhomogeneous Second Order ODE

[0131413]

Homework Statement

What is the value of a such that the solution of the initial-value problem satisfies lim_x->infinity y(x) = 0?

y''+y'=e^(-x), y(0)=1, y'(0)=a

Homework Equations

The Attempt at a Solution

Not sure what to do with the missing y term...

y_p=Ae^(-x), y'_p=-A^(-x), y''_p=A^(-x)
Ae^(-x)-A^(-x) = 0 so

y_p=Axe^(-x), y'_p=-Axe^(-x)+Ae^(-x). y''_p=Axe^(-x)-Ae^(-x)-Ae^(-x)
Axe^(-x)-Ae^(-x)-Ae^(-x)-Axe^(-x)+Ae^(-x)=e^(-x)
-Ae^(-x)=e^(-x)
A=-1

The general solution is C₁e^(0)+C₂e^(-x) (?)

y=te^(-x)+C₁e^(0)+C₂e^(-x)
y'=-te^(-x)+e^(-x)-C₂e^(-x)

y(0)=0+C₁-C₂=1
y'(0)=0+1-C₂=a

C₂=1-a

I know this is wrong, multiple numbers can fit into a. Thanks!

 

[LCKurtz]

Homework Statement

What is the value of a such that the solution of the initial-value problem satisfies lim_x->infinity y(x) = 0?

y''+y'=e^(-x), y(0)=1, y'(0)=a

Homework Equations

The Attempt at a Solution

Not sure what to do with the missing y term...

y_p=Ae^(-x), y'_p=-A^(-x), y''_p=A^(-x)
Ae^(-x)-A^(-x) = 0 so

y_p=Axe^(-x), y'_p=-Axe^(-x)+Ae^(-x). y''_p=Axe^(-x)-Ae^(-x)-Ae^(-x)
Axe^(-x)-Ae^(-x)-Ae^(-x)-Axe^(-x)+Ae^(-x)=e^(-x)
-Ae^(-x)=e^(-x)
A=-1

The general solution is C₁e^(0)+C₂e^(-x) (?)

That's the general solution (complementary solution) of the homogeneous equation, yes. No point in the e⁰ term; it is just 1:
y_c = C₁ + C₂e^-x.

But up above you found the particular solution. Since A = -1, what do you get for y_p?

y=te^(-x)+C₁e^(0)+C₂e^(-x)

There shouldn't be any t in this equation. What is the correct y_p?

y'=-te^(-x)+e^(-x)-C₂e^(-x)

y(0)=0+C₁-C₂=1
y'(0)=0+1-C₂=a

C₂=1-a

I know this is wrong, multiple numbers can fit into a. Thanks!

Yes, it's wrong but that isn't why. Remember, changing a changes the problem so different a gives the answer to a different equation. Fix your y and express your answer in terms of a. And lose the variable t.

 

[0131413]

I'm sorry that t is supposed to be an x. I did forget the negative sign when trying to solve the general solution, ouch.

I thought it would be xe^(-x) since using y_p=Ae^(-x) I ended up with Ae^(-x)-Ae^(-x).

Now I have

y=-xe^(-x)+C₁+C₂e^(-x)
y'=xe^(-x)-e^(-x)-C₂e^(-x)

y(0)=C₁+C₂=1
y'(0)=-1-C₂=a

C₁=1+1+a=0 (Since C₁ needs to be 0 for x->infinity to be 0)

So a is -2?

 

[LCKurtz]

I agree. So what is your final solution to the problem? Once you have it you can plug it into the equation, initial conditions and check the limit and you will know for sure you have no arithmetic mistakes.

 

[0131413]

I agree. So what is your final solution to the problem? Once you have it you can plug it into the equation, initial conditions and check the limit and you will know for sure you have no arithmetic mistakes.

C₂=-3

The equation that satisfies the problem is y=-xe^(-x)-3e^(-x)

y'=xe^(-x)-e^(-x)+3e^(-x)
y''=-xe^(-x)+e^(-x)+e^(-x)-3e^(-x)

-xe^(-x)+2e^(-x)-3e^(-x)+xe^(-x)-e^(-x)+3e^(-x)=e^(-x)

Final answer to the question of what value of a satisfies the problem:

a=-2

Thanks a lot. :)