# solve a non homogeneous differential equation

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I need a hint on how to solve the following DE.

x'' = ex -1.

The general answer to the equation will be the sum of inhom. + hom. solutions of the equation.

Any help is appreciated.

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• 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?

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$A\cdot e^x - \dfrac {1}{2} x^2$ (right?)

however, considering the general form :

$x'' + ax' + bx = g(y)$

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 by allen_83
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I need a hint on how to solve the following DE.

x'' = ex -1.

The general answer to the equation will be the sum of inhom. + hom. solutions of the equation.

Any help is appreciated.

Integrate twice.

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Integrate twice.
I'm assuming you meant separation of variables, and then integrating twice, right?
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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.

$\dfrac {d^2 \ x}{dt^2} - e^x = -1$

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 $\dfrac {d^2 \ x}{dt^2} = e^t -1$ ?

If so, then integrate twice.

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Sorry, brain fart.

None of the techniques discussed are going to work, because this ODE is non-linear.

$\dfrac {d^2 \ x}{dt^2} - e^x = -1$

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 $\dfrac {d^2 \ x}{dt^2} = e^t -1$ ?

If so, then integrate twice.

That's right. This ODE is non-linear.

The problem is as it is.

$x''(t) = e^{ x(t)} - 1 for x'(0)= 0 and x(0) = 1$

Attempts:

$x''(t) + 1 = e^{x(t)}; \lambda^2+1 = 0 \Rightarrow \lambda_{1} = i, \lambda_{2} = -i ; x(t) = D_{1}cos(t) + D_{2}sin(t) \because x(0) = 1 , x'(0) = 0$

solvin for D1 and D2 :

D1 = 1, D2 = 0

and pluging them back in :

$x(t) = cos(t)$

$x_{nonhom}(t) = A\cdot e^x - \dfrac{x^2}{2}$

$x'_{nonhom}(t) = A\cdot e^x - x$

$x''_{nonhom}(t) = A\cdot e^x - 1$

$A\cdot e^x - 1 + 1 = e^x \Rightarrow A = 1$

$x_{nonhom}(t) = e^x - \dfrac{x^2}{2}$

$x = x_{nonhom}(t) + x_{hom}(t)$

$x = e^{x(t)} - \dfrac{x(t)^2}{2} + cos(t)$

correct me if I'm wrong.

Edited by allen_83
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That's right. This ODE is non-linear.

The problem is as it is.

$x''(t) = e^{ x(t)} - 1 for x'(0)= 0 and x(0) = 1$

Attempts:

$x''(t) + 1 = e^{x(t)}; \lambda^2+1 = 0 \Rightarrow \lambda_{1} = i, \lambda_{2} = -i ; x(t) = D_{1}cos(t) + D_{2}sin(t) \because x(0) = 1 , x'(0) = 0$

solvin for D1 and D2 :

D1 = 1, D2 = 0

and pluging them back in :

$x(t) = cos(t)$

$x_{Inhom}(t) = A\cdot e^x - \dfrac{x^2}{2}$

$x'_{Inhom}(t) = A\cdot e^x - x$

$x''_{Inhom}(t) = A\cdot e^x - 1$

$A\cdot e^x - 1 + 1 = e^x \Rightarrow A = 1$

$x_{Inhom}(t) = e^x - \dfrac{x^2}{2}$

$x(t) = x_{inhom}(t) + x_{hom}(t)$

$x(t) = e^x - \dfrac{x^2}{2} + cos(t)$

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. $x(t) = e^x - \dfrac{x^2}{2} + cos(t)$ fails to express x explicitly as a function of t.

3. $\dfrac {d}{dt}( e^x - \dfrac{x^2}{2} + cos(t)) = e^{x(t)} \frac {dx}{dt} -x \frac {dx}{dt} - sin(t)$

$\dfrac {d^2}{dt^2}( e^x - \dfrac{x^2}{2} + cos(t))$$= 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)$ This does not appear to help very much.

4. The technique you are applying in your

$x''(t) + 1 = e^{x(t)}; \lambda^2+1 = 0 \Rightarrow \lambda_{1} = i, \lambda_{2} = -i ; x(t) = D_{1}cos(t) + D_{2}sin(t) \because x(0) = 1 , x'(0) = 0$

is intended for homogeneous linear equations. But this is all screwed up. First because $x''+1=0$ is not homogeneous, and second because it does not help with $x''(t) + 1 = e^{x(t)}$

The solutions to $x"=-1$ are $x(t) = - \frac {t^2}{2} - at + c$, so you can see that you have misapplied the characteristic equation.

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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 : $x'' = f(t,x,x')$ with both x and t missing if I were to think of it as a non-linear DE.

$x(t) = - \frac {t^2}{2} - at + c$

how ?

Edited by allen_83

how ?

integrate twice

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