# Hard D.E

1. Dec 11, 2005

(t*y'(t))' = t*Dirac(t-2), t>0
we know that y(t) = 1 lim t--> o+
solve for y(t)

the solution says that by using
f(t)*Dirac_a(t) = f(a) * Dirac_a(t) we get

(t*y'(t))' = 2*Dirac(t-2) Dont see how they get that

2. Dec 11, 2005

### CarlB

Ah, by "Dirac" you mean the Dirac delta function, not the Dirac operator.

Yes, the second equation is applied to the first to convert the "t" to the value "2". Another way of putting this.

Since Dirac(t-2) is zero except when t=2, you might as well go replace t with 2 everywhere that is multplied by Dirac(t-2).

Hope this helps.

Carl

3. Dec 12, 2005

### saltydog

You know Larsson, I'd like to know what the solution is to this problem. Me, I'd rewrite it as:

$$ty^{''}+y^{'}=t\delta_2(t)$$

Taking the Laplace transform of both sides results in:

$$-sY-s^2Y^{'}=2e^{-2s}$$

Solving for Y I get:

$$Y(s)=-\frac{2}{s}\text{Ei}(-2s)+\frac{c}{s}$$

where Ei[x] is the exponential integral function.

Inverting I get:

$$y(t)=2\text{ln}(t/2)\text{UnitStep}(t-2)+c$$

You said the limit of y(t) as t goes to zero is 1 so I'd let c=1. However, I'm not very confident of this as the question of initial conditions for both the DE in t and the DE in s is not well defined in my mind. What is your solution?

4. Dec 12, 2005

Almost correct. + A*ln(t) and you're right

While we're at it I have another:

find a function y which you can apply to

y'(t) - int(2*sin(t - tau) * y(tau) d tau tau = 0 .. t) = cos(t) y(0) = -1/2

Bt using the definition of convolution I get that it can be written

y'(t) - 2 sin(t)* y(t) = cos(t)

laplacetransformation gives

sY +1/2 - 2/(s^2+1) * Y = s/(s^2+1)

Y (s - 2/(s^2+1)) = s/(s^2+1) - 1/2

Y ( s(s^2+1) - 2) = s - (s^2 + 1)/2

Y = s/(s(s^2+1) - 2) - (s^2 + 1)/(2*(s^2 + 1))

and from here everything just crashes, dont get anything that makes sence

5. Dec 12, 2005

### saltydog

Thanks, I'll spend more time with it as I think it's an interesting problem.

Looks like:

$$y^{'}-2\int_0^t Sin[t-\tau]y(\tau)d\tau=Cos[t],\quad y(0)=1/2$$

That too is very interesting. I'll work with it too.

6. Dec 13, 2005

### CarlB

An obvious solution to the homogeneous problem is

$$y(t) = c \; e^{-2\cos(t)}$$

One of my old physics professors would say that a solution to the inhomogeneous problem can be obtained by a "simple quadrature".

Carl

Last edited: Dec 13, 2005