allen_83

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Posts posted by allen_83


Sorry, brain fart.
None of the techniques discussed are going to work, because this ODE is nonlinear.
[math] \dfrac {d^2 \ x}{dt^2}  e^x = 1[/math]
Given the OP's statement that a solution in terms of a general solution to the homogeneous equation plus a particular solution to the inhomogeneous equation is expected (which applies to a linear ODE), I suspect that the equation written is not the equation of the real problem.
Should it be [math] \dfrac {d^2 \ x}{dt^2} = e^t 1[/math] ?
If so, then integrate twice.
That's right. This ODE is nonlinear.
The problem is as it is.
[math] x''(t) = e^{ x(t)}  1 for x'(0)= 0 and x(0) = 1 [/math]
Attempts:
[math]x''(t) + 1 = e^{x(t)}; \lambda^2+1 = 0 \Rightarrow \lambda_{1} = i, \lambda_{2} = i ; x(t) = D_{1}cos(t) + D_{2}sin(t) \because x(0) = 1 , x'(0) = 0 [/math]
solvin for D1 and D2 :
D1 = 1, D2 = 0
and pluging them back in :
[math] x(t) = cos(t)[/math]
[math]x_{nonhom}(t) = A\cdot e^x  \dfrac{x^2}{2}[/math]
[math] x'_{nonhom}(t) = A\cdot e^x  x[/math]
[math]x''_{nonhom}(t) = A\cdot e^x  1 [/math]
[math]A\cdot e^x  1 + 1 = e^x \Rightarrow A = 1[/math]
[math]x_{nonhom}(t) = e^x  \dfrac{x^2}{2}[/math]
[math] x = x_{nonhom}(t) + x_{hom}(t)[/math]
[math] x = e^{x(t)}  \dfrac{x(t)^2}{2} + cos(t) [/math]
correct me if I'm wrong.
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[math]A\cdot e^x  \dfrac {1}{2} x^2[/math] (right?)
however, considering the general form :
[math]x'' + ax' + bx = g(y)[/math]
there is no place for a constant c. So I wonder if I'm actually allowed to move 1 to the left hand side of the equation(?)
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I need a hint on how to solve the following DE.
x'' = e^{x }1.
The general answer to the equation will be the sum of inhom. + hom. solutions of the equation.
Any help is appreciated.
0
solve a non homogeneous differential equation
in Analysis and Calculus
Posted · Edited by allen_83
alright, wanted to give "the tree"'s idea a try.
got no clue. am stuck.
this is a type : [math] x'' = f(t,x,x') [/math] with both x and t missing if I were to think of it as a nonlinear DE.
how ?
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