# Hard diff-eq

1. Oct 31, 2004

### hedlund

I'm finding this diff equation hard
$$y'' + \ln{y} = yx$$
How do I solve it?

2. Oct 31, 2004

### James R

Numerically.

3. Oct 31, 2004

### PerennialII

Willing to agree with the previous comment .... I'm not sure how well the classical numerical solution methods bite, but if nothing else you can discretize and do for example a simplistic DM or FE etc. analysis.

4. Oct 31, 2004

### hedlund

I don't want it numerically, I want it exact if possible

5. Oct 31, 2004

### Dr Transport

rearrange: $$y'' - xy = -\ln(y)$$, solve the homogeneous equation for $$y$$ then use those solutions as an integrating factor, or Green's function to solve the equation.

dt

6. Nov 1, 2004

### HallsofIvy

No, that won't work: this is not a linear equation. The right hand side is ln(y), NOT ln(x)!

7. Nov 1, 2004

### arildno

We MIGHT make some headway if we seek an asymptotic approximation in the vicinity of x=0 (i.e, we make a supposition that our unknown function y(x) is defined at x=0)
Let us state an initial condition:
$$y(0)=y_{0}>0$$
$$y'(0)=y'_{0}$$

Define:
$$y(x)=Y(x)+y_{0}$$,
so that:
$$Y(0)=0,Y'(0)=y_{0}'$$

In the vicinity of x=0, we have:
$$ln(y)=ln(y_{0}+Y)=ln(y_{0})+ln(1+\frac{Y}{y_{0}})\approx{ln(y_{0})}+\frac{Y}{y_{0}}$$
$$yx=y_{0}x+Yx\approx{y}_{0}x$$

Hence, close to x=0, we have the differential equation in Y:
$$Y''+\frac{Y}{y_{0}}=y_{0}x-ln(y_{0}), Y(0)=0 (1)$$
The general solution of the homogenous equation (that is, $$Y''+\frac{Y}{y_{0}}=0$$) is:
$$Y_{h}(x)=A\cos(\frac{x}{\sqrt{y_{0}}})+B\sin(\frac{x}{\sqrt{y_{0}}})$$
A particular solution to (1) is the linear function:
$$Y_{p}=y_{0}^{2}x-y_{0}ln(y_{0})$$
We therefore set
$$Y(x)=Y_{h}+Y_{p}$$
$$Y(0)=0\to{A}=y_{0}ln(y_{0})$$
Whereas:
$$Y'(0)=y_{0}'\to{B}=y_{0}'\sqrt{y_{0}}-(y_{0})^{\frac{5}{2}}$$

Hence, we get the asymptotic solution, to first order:
$$y(x)=y_{0}+y_{0}ln(y_{0})\cos(\frac{x}{\sqrt{y_{0}}})+(y_{0}'\sqrt{y_{0}}-(y_{0})^{\frac{5}{2}})\sin(\frac{x}{\sqrt{y_{0}}})+y_{0}^{2}x-y_{0}ln(y_{0})$$

I would like to emphasize that this is only a first order approximation, valid in the limit $$x\to0$$

Last edited: Nov 1, 2004
8. Dec 4, 2004

### hedlund

(1) y'' + ln(y) = yx
(2) ln(y'' + ln(y)) = ln(yx) = ln(y)+ln(x)
(3) ln(y'' + ln(y)) = ln(y) + ln(x)
(4) ln(y'' + ln(y)) - ln(y) = ln(x)
(5) ln(y''/ln(y) + ln(y)/ln(y)) = ln(x)
(6) ln(y''/ln(y) + 1) = ln(x)
(7) y''/ln(y) + 1 = x
(8) y'' + ln(y) = ln(y)*x

Since the left side of (8) is equal to the left hand side of (1) this must give that ln(y) * x = y * x => ln(y) = y => y = e^y.

However I don't think there exist a solution to y = e^y ... if it exists it must a complex integer so y is a constant function. Using math software this gives y = 0.3181315052047641 - 1.3372357014306895i

9. Dec 4, 2004

### arildno

Step 5) is wrong hedlund
From 4), we have:
ln(y''+ln(y))-ln(y)=ln(y''/y+ln(y)/y)
As I'm sure you agree with..

10. Dec 4, 2004

### hedlund

Yeah it was plain stupid ... I thought of it as ln(ln(y)) :/

11. Dec 4, 2004

### arildno

To able to find exact solutions to (non-linear) diff.eqs are more often than not just a happy chance, NEVER the rule.
In fact, you'll meet just about every exact solution of diff.eqs which has been /can be found through your studies, they are extremely few in number.