I know how to solve a linear differential equation. But question is that what does that mean transient terms in general solution.

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    $\begingroup$ when $t$ or $x\rightarrow \infty$ the transient components vanish and only steady state long term effects are seen in (the short range time terms vanish) the solution. Please give your specific example. $\endgroup$ – Narasimham Feb 27 '18 at 11:02
  • $\begingroup$ x^2 dy/dx + x(x+2)y = e^x the solution of this Linear Differential Equation is y = 1/2 x^-2 e^x + Cx^-2 e^-x so identify transient terms??? Is x = 0 a singular point??? $\endgroup$ – Naeem Ivy Feb 27 '18 at 15:03

The general solution $x(t)$ of a scalar linear differential equation


can be expressed as a sum of the homogeneous solution $x_\text{h}(t)$ particular solution $x_\text{p}(t)$


The homogeneous solution is sometimes referred as the natural solution, unforced solution (which means $u(t)\equiv 0$) or transient solution.

If the differential equation is stable, which is equivalent to the statement that all the eigenvalues (roots of the characteristic equation) have a strictly negative real part, then the transient solution will be insignificant for $t\to \infty$, as $x_\text{h}(t) \to 0$ and the behaviour of the system is dominated by the particular solution $x_\text{p}(t)$.

  • $\begingroup$ Thanks bro 👍👍👍👍 $\endgroup$ – Naeem Ivy Feb 27 '18 at 16:47

If solution of

$$ x^2 dy/dx + x(x+2)y = e^x $$


$$ y = \frac{e^x}{ 2 x^2} +\frac{ C}{x^2} -\frac{2}{e^{x}} $$

then the last two terms tend to zero, whatever (even infinite) starting value may be.

  • $\begingroup$ So nice of . . ❤❤❤ excellent $\endgroup$ – Naeem Ivy Feb 27 '18 at 21:01

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