allen_83 Posted March 21, 2011 Share Posted March 21, 2011 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. Link to comment Share on other sites More sharing options...

the tree Posted March 21, 2011 Share Posted March 21, 2011 Actually the general sollution is the sum of the homogeneous and implicit solutions. Rearranging it like so will make it a little easier: [imath]x'' +1 = e^x[/imath] You should be able to solve the homogeneous equation [imath]x'' + 1=0[/imath] without much difficulty. Finding an implicit solution is essentially just well informed guesswork - what do you think the solution will look like? Link to comment Share on other sites More sharing options...

allen_83 Posted March 21, 2011 Author Share Posted March 21, 2011 (edited) [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(?) Edited March 21, 2011 by allen_83 Link to comment Share on other sites More sharing options...

DrRocket Posted March 22, 2011 Share Posted March 22, 2011 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. Integrate twice. Link to comment Share on other sites More sharing options...

the tree Posted March 22, 2011 Share Posted March 22, 2011 Integrate twice.I'm assuming you meant separation of variables, and then integrating twice, right? Link to comment Share on other sites More sharing options...

DrRocket Posted March 22, 2011 Share Posted March 22, 2011 I'm assuming you meant separation of variables, and then integrating twice, right? Sorry, brain fart. None of the techniques discussed are going to work, because this ODE is non-linear. [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. Link to comment Share on other sites More sharing options...

allen_83 Posted March 22, 2011 Author Share Posted March 22, 2011 (edited) Sorry, brain fart. None of the techniques discussed are going to work, because this ODE is non-linear. [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 non-linear. 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. Edited March 22, 2011 by allen_83 Link to comment Share on other sites More sharing options...

DrRocket Posted March 22, 2011 Share Posted March 22, 2011 That's right. This ODE is non-linear. 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_{Inhom}(t) = A\cdot e^x - \dfrac{x^2}{2}[/math] [math] x'_{Inhom}(t) = A\cdot e^x - x[/math] [math]x''_{Inhom}(t) = A\cdot e^x - 1 [/math] [math]A\cdot e^x - 1 + 1 = e^x \Rightarrow A = 1[/math] [math]x_{Inhom}(t) = e^x - \dfrac{x^2}{2}[/math] [math] x(t) = x_{inhom}(t) + x_{hom}(t)[/math] [math] x(t) = e^x - \dfrac{x^2}{2} + cos(t) [/math] correct me if I'm wrong. 1. Since the equation is non-linear, what makes you think that the solution is expressible as the sum of "the general homogeneous solution" plus an "inhomogeneous solution" ? When the differential operator is linear, any two solutions of the inhomogeneous equation differ by a solution of the homogeneous equation. Not so for non-linear equations. 2. [math] x(t) = e^x - \dfrac{x^2}{2} + cos(t) [/math] fails to express x explicitly as a function of t. 3. [math] \dfrac {d}{dt}( e^x - \dfrac{x^2}{2} + cos(t)) = e^{x(t)} \frac {dx}{dt} -x \frac {dx}{dt} - sin(t) [/math] [math] \dfrac {d^2}{dt^2}( e^x - \dfrac{x^2}{2} + cos(t))[/math][math] = e^{x(t)} (\frac {dx}{dt})^2 + e^{x(t)} \frac {d^2x}{dt^2} - (\frac {dx}{dt})^2 -x \frac {d^2x}{dt^2} - cos(t) [/math] This does not appear to help very much. 4. The technique you are applying in your [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] is intended for homogeneous linear equations. But this is all screwed up. First because [math]x''+1=0[/math] is not homogeneous, and second because it does not help with [math]x''(t) + 1 = e^{x(t)}[/math] The solutions to [math]x"=-1[/math] are [math] x(t) = - \frac {t^2}{2} - at + c [/math], so you can see that you have misapplied the characteristic equation. Link to comment Share on other sites More sharing options...

allen_83 Posted March 23, 2011 Author Share Posted March 23, 2011 (edited) 1. Since the equation is non-linear, what makes you think that the solution is expressible as the sum of "the general homogeneous solution" plus an "inhomogeneous solution" ? When the differential operator is linear, any two solutions of the inhomogeneous equation differ by a solution of the homogeneous equation. Not so for non-linear equations. 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 non-linear DE. [math] x(t) = - \frac {t^2}{2} - at + c [/math] how ? ^{} Edited March 23, 2011 by allen_83 Link to comment Share on other sites More sharing options...

DrRocket Posted March 23, 2011 Share Posted March 23, 2011 how ? integrate twice Link to comment Share on other sites More sharing options...

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