Solving Diff Equation: y'' + ln(y) = yx

[hedlund]

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

 

[James R]

Numerically.

 

[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.

 

[hedlund]

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

 

[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

 

[HallsofIvy]

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

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

 

[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
In addition, we set:
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

 

[hedlund]

I've thought about this a lot ...

(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

 

[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..

 

[hedlund]

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..

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

 

[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.