# Find solution with initial value

1. Apr 27, 2010

### invisible_man

y'' - ty' + y = 1, y(0) = 1, y'(0) =2

I'm trying to solve by Laplace Transform and I got stuck at the end.Anyone can help me to solve it?

2. Apr 27, 2010

### matematikawan

Did you compute the Laplace transform of ty'(t) correctly? I think the Laplace transform of this term and computing the Laplace inversion later are the only diffuculty that I foresee.

3. Apr 27, 2010

### invisible_man

The Laplace Transform of ty' = -L[y]'

4. Apr 27, 2010

### matematikawan

I think so. Except that the RHS differentiation is wrt s, i.e. $$-\frac{dF}{ds}= -\frac{d}{ds}(sY-1)$$

Last edited: Apr 28, 2010
5. Apr 28, 2010

### kosovtsov

Your ODE is not an ODE with constant coefficients, so it can not be solved by Laplace Transform.

An integrating factor to your ODE

$$\mu=t\exp(-\frac{t^2}{2})$$

and corresponding first integral is

$$I=\exp(-\frac{t^2}{2})(t\frac {d y}{d t}-y+1)+C1$$

in sense that

$$\frac {d I}{d t}=t\exp(-\frac{t^2}{2})(\frac {d^2 y}{d t^2}-t\frac {d y}{d t}+y-1),$$

what is equivalent to your ODE.

Substituting $$t=0$$ in $$I$$ we get

$$-y(0)+1=C1$$

and as $$y(0)=1$$ then

$$C1=0.$$

Now you have to solve first order ODE ($$I=0$$)

$$t\frac {d y}{d t}-y(t)+1=0.$$

Its solution is

$$y(t) = tC2+1$$

$$y(t) = 2t+1$$

6. Apr 28, 2010

### matematikawan

Some linear DE with variable coefficients can be solve via Laplace transform. It all depends on whether we can solve for Y(s) and later invert it. In this particular case, the transformed equation will be a linear first order DE in Y(s) which I think can be solve for Y(s) quite easily. But whether it can be invert easily, need to look at the expression first.

7. Apr 28, 2010

### kosovtsov

Can you give an example, please?

8. Apr 28, 2010

### matematikawan

I remember some time ago somebody try to solve Bessel equation via Laplace transform.
But that person didn't proceed to see them through.

But for the Bessel equation of order zero, it can be done.
x2y''+xy'+x2y = 0
xy''+y'+xy = 0

Apply Laplace transform and solve for Y(s)
$$-\frac{d}{ds}(s^2Y-as-b) + sY-a-\frac{dY}{ds}$$

$$\frac{dY}{ds}=-\frac{sY}{s^2+1}$$

$$Y(s)=\frac{A}{\sqrt{s^2+1}}$$

The solution will be the inversion of Y(s).

Last edited by a moderator: Apr 25, 2017