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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$\begingroup$The general solution $x(t)$ of a scalar linear differential equation
$$x^{(n)}(t)+a_{n-1}x^{(n-1)}(t)+...+a_1x'(t)+a_0x(t)=u(t)$$
can be expressed as a sum of the homogeneous solution $x_\text{h}(t)$ particular solution $x_\text{p}(t)$
$$x(t)=x_\text{h}(t)+x_\text{p}(t).$$
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 $x_\text{p}(t)$.
$\endgroup$ 1 $\begingroup$If solution of
$$ x^2 dy/dx + x(x+2)y = e^x $$
is
$$ y = \frac{e^x}{ 2 x^2} +\frac{ C}{x^2} -\frac{2}{e^{x}} $$
then the last two terms tend to zero, whatever (even infinite) starting value may be.
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