Answer:
Step-by-step explanation:
Given that the differential equation is:
[tex]5 \dfrac{dy}{dx}+ 45 y = 9\\[/tex]
by dividing both sides by 5; we have:
[tex]\dfrac{dy}{dx}+ 9y = \dfrac{9}{5} --- (1)[/tex]
which is a form of a linear equation.
The integrating factor is:
[tex]\mu = e^{\int}^{9dx }[/tex]
[tex]\mu = e^{9x }[/tex]
multiplying eqution (1) with [tex]\mu = e^{9x }[/tex]
[tex]e^{9x}(\dfrac{dy}{dx}+ 9y)= (\dfrac{9}{5})e^{9x}[/tex]
[tex]\dfrac{d}{dx}(ye^{9x})= (\dfrac{9}{5})e^{9x}[/tex]
Using integration on both sides.
[tex]\int \dfrac{d}{dx}(ye^{9x}) dx= \int \dfrac{9}{5} e^{9x} \ dx[/tex]
[tex]ye^{9x}=\dfrac{9}{5} \dfrac{e^{9x}}{9}+C[/tex]
[tex]ye^{9x}= \dfrac{e^{9x}}{5}+C[/tex]
[tex]y= \dfrac{1}{5}+Ce^{-9x}[/tex]
Thus, the general solution is : [tex]y= \dfrac{1}{5}+Ce^{-9x} ; -\infty < x < \infty[/tex]
In general solution:
[tex]y_c = ce^{-9x} \ and \ y_p = \dfrac{1}{5}[/tex]
To talk of larger values of x, the value of [tex]y_c(x)[/tex] is said to be negligible.
This implies that:
[tex]y_c \to 0 \ as \ x \to \infty[/tex]
Thus,
[tex]y_c (x) = ce ^{-9x}[/tex] signifies a transient term in the general solution.
