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I have a very complicated differential equation that can not be solved analytically and I show a on the simple example:

Example 1: $$y''(x)-y(x)=0\tag{1}$$ with boundary conditions: $$y(1)=1,y'(2)=1\tag{2}$$ solution is: $$y(x)=\frac{e^{-x-1} \left(e^{2 x}+e^{2 x+1}-e^3+e^4\right)}{1+e^2}\tag{3}$$ then using the substitution of $$v(x)=\frac{y(x)}{x}$$ I have a NEW equation: $$x v''(x)+2 v'(x)-x v(x)=0\tag{4}$$ How to find the boundary conditions of the NEW differential equation? $$v(?)=?,v'(?)=?$$ EDITED! $$v(x)= \frac{y(x)}{x}\tag{5}$$ $$v(1)=\frac{y(1)}{1}=\frac{1}{1}=1$$ $$v'(x)= \frac{y'(x)x-y(x)}{x^2}=\frac{y'(2)*2-y(x)}{2^2}= \frac{1*2-y(x)}{2^2}$$ I don't have a $y(2)=?$ Assume $y(x)$ is $y(1)=1$ then: $$v'(2)= \frac{1*2-1}{2^2}=1/4$$ The new boundary conditions are: $$v(1)=1,v'(2)=1/4\tag{6}$$Solution with New equation$(4)$ and new boundary conditions $(6)$ is: $$v(x)=\frac{e^{-x-1} \left(3 e^{2 x}+e^{2 x+1}-e^3+e^4\right)}{\left(3+e^2\right) x}$$ then we substitute to $(5)$ and check solutions is equal: $$\frac{e^{-x-1} \left(3 e^{2 x}+e^{2 x+1}-e^3+e^4\right)}{\left(3+e^2\right)}\neq\frac{e^{-x-1} \left(e^{2 x}+e^{2 x+1}-e^3+e^4\right)}{1+e^2}$$ Is NOT !!.

Example 2: $$y''(x)-y(x)=0$$ with boundary conditions: $$y'(1)=1,y'(2)=1$$
How to find the boundary conditions of the NEW differential equation?

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    $\begingroup$ You can just compute $v(1)$ and $v'(2)$ directly.... $\endgroup$ – user296602 Jan 18 '16 at 17:43
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Hint: we have : $$ v(x)=\frac{y(x)}{x} $$ than we find : $$ v'(x)= \frac{d}{dx}v(x)=\frac{d}{dx}\frac{y(x)}{x}= \frac{y'(x)x-y(x)}{x^2} $$ now put $x=1$ and $x=2$ and use the given initial conditions for $y(1)$ and $y'(2) $.


The starting equation $y''(x)-y(x)=0$ has solution: $y=ae^{-x}+be^x$.

From the boundary condition $y(1)=1$ we can find: $$ \frac{a}{e}+eb=1 \Rightarrow a=e(1-eb) $$ For the derivative $y'=-ae^{-x}+be^x $ the other condition $y'(2)=1$ gives: $$ \frac{-a}{e^2}+e^2b=1 $$ and substituting the value of $a$ we have: $$ -1+eb+e^3b=e \Rightarrow b=\frac{e+1}{e(e^2+1)} $$ and $$ a=\frac{e^2(e-1)}{e^2+1} $$ and this solve the problem. But you want solve the same problem with the substitution $v(x)=\frac{y(x)}{x}$. So the equation becomes: $$ xv''(x)+2v'(x)-xv(x)=0 $$ that has obviously the solution: $$ v(x)=\frac{ae^{-x}}{x}+\frac{be^{x}}{x} $$ Now we use the new conditions to find the constants $a,b$.

From $v(1)=\frac{y(1)}{1}=1$ we find: $$ a=e(1-eb) $$ so: $$ (1)\qquad \quad v(x)=\frac{e(1-eb)e^{-x}}{x}+\frac{be^x}{x} $$ and $$ v'(x)= \frac{-e(1-eb)xe^{-x}-e(1-eb)e^{-x}}{x^2}+\frac{bxe^x-be^x}{x^2} $$ so: $$ (2) \qquad \quad v'(2)=\frac{3(eb-1)}{4e} $$ Now use the condition for $v'(2)$ noting that $y(x)=xv(x)$ $$ v'(2)=\frac{2\cdot y'(2)-2\cdot v(2)}{4}=\frac{1-v(2)}{2} $$ Now we can substitute $v(2)$ from the equation $(1)$ and, with a bit of algebra, we find: $$ v(2)=\frac{2e-1+eb-e^3b}{4e} $$ Equating with $(2)$ we finally find the value of $b$: $$ b=\frac{e+1}{e(e^2+1)} $$ so you see that the constants are the same in the two functions.

(I hope that i've no typos !)

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  • $\begingroup$ This is a wrong answer! $\endgroup$ – Mariusz Iwaniuk Jan 18 '16 at 19:50
  • $\begingroup$ I'm interesting to know why it's a wrong answer. Can you motivate? $\endgroup$ – Emilio Novati Jan 18 '16 at 22:03
  • $\begingroup$ Yours hint is very good but not solve my problem.First boundary conditions v(1)=1 is ok,a second condition v'(2)=1/4 is not. $\endgroup$ – Mariusz Iwaniuk Jan 19 '16 at 11:51
  • $\begingroup$ I've added a complete solution to my answer. I hope it's useful :) $\endgroup$ – Emilio Novati Jan 19 '16 at 15:44
  • $\begingroup$ Assume if I have different differential equation and I can't find analyticall solution to this equation,It also will I not find a new boundary conditions of the NEW differential equation? $\endgroup$ – Mariusz Iwaniuk Jan 20 '16 at 10:51

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