# allen_83

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## Posts posted by allen_83

### solve a non homogeneous differential equation

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 ?

### solve a non homogeneous differential equation

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.

### solve a non homogeneous differential equation

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

### solve a non homogeneous differential equation

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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