Solve second order diff equation using substitution

  • Thread starter Gekko
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  • #1
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Homework Statement



d2y/dx2-dy/dx+y*exp(2x) = x*exp(2x)-1

substitute t=exp(x) and set z(t)=y(x) and rewrite hence find all solutions

The Attempt at a Solution



Rewriting gives:
d2z/dt2-dz/dt+z*t^2=(ln(t) * t^2) - 1

however I dont see how this in any way helps us...
 

Answers and Replies

  • #2
HallsofIvy
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You haven't changed the derivatives properly. You cannot just replace "dy/dx" with "dy/dt". You have to use the chain rule.

dy/dx= (dy/dt)(dt/dx) so if t= e^x, dy/dx= e^x(dy/dt)= t(dy/dt).

d^2y/dx^2= (d/dx)(dy/dx)= d/dx(t dy/dt)= t(d/dt(t(dy/dt)). By the product rule, that is t^2 d^2y/dt^2+ t dy/dt.
The equation becomes (t^2 d^2y/dt^2+ t dy/dt)- t dy/dt+ t^2 y= t^2(d^2y/dt^2+ y)= t^2ln(t)- 1 so that d^2y/dt^2+ y= ln(t)- t^{-2}.

The related homogeneous equation is easy to solve and I would recommend "variation of parameters" to find a specific solution to the entire equation.
 
  • #3
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first of all:
"d2y/dx2" does not equals "d2z/dt2" and "dy/dx" does not equals "dz/dt".

but you right in saying that exp(2x) = t^2 and x = ln(t)

you need to use the "chain rule" in differentiation !

t=e^x

z(t)=y(x)

dz(t)/dx = dy(x)/dx => (dz(t)/dt)*(dt/dx) = y'(x)

*try to go from here :)
*don't forget the differentiate correctly when you do d^2z(t)/dx^2 = d^2y(x)/dx^2, it has the form (u*v)' = u'*v + u*v'

Hope it will lead you to the right answer ! :)
 
  • #4
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HallsofIvy beat me to it :)

also I love Lagrange's variation of parameters !

Here is how you do it:

(( in your case y=z, x=t ))

y1(x), y2(x) : are two linearly independent homogeneous solutions.
g(x) : the particular expression, in your case = ln(t)- t^{-2}.

u'1(x)*y1(x) + u'2(x)*y2(x) = 0
u'1'(x)*y'1(x) + u'2(x)*y'2(x) = g(x)

solve the system of equations and find u1(x) and u2(x) by integration.
the particular part of the answer will be:

Y(x) = u1(x)*y1(x) + u2(x)*y2(x).

the complete answer for the ODE (A,B are constants):

y(x) = [A*y1(x)] + [B*y1(x)] + [u1(x)*y1(x) + u2(x)*y2(x)].

*Lagrange was a genius !
 

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