Subsection 4.1.1 First Order Linear Differential Equation
The first order linear differential equation is obtained by putting \(n = 1\) in the standard form of linear equation (4.0.1), i.e.,
\begin{equation*}
f_{o}(x)\frac{\,dy}{\,dx}+ f_{o}(x)y =F(x)
\end{equation*}
or,
\begin{equation}
\frac{\,dy}{\,dx}+P(x)y = Q(x)\tag{4.1.6}
\end{equation}
where \(P(x)\) and \(Q(x)\) are functions of \(x\) or constants. To find its solution, first consider the homogeneous equation by setting \(Q(x) = 0\text{.}\)
\begin{equation*}
\frac{\,dy}{\,dx}+P(x)y = 0
\end{equation*}
\begin{equation*}
\frac{1}{y}\frac{\,dy}{\,dx} =-P(x)y
\end{equation*}
integrating, we get -
\begin{equation*}
\log y =-\int P(x)\,dx +C
\end{equation*}
where \(C\) is an arbitrary constant of integration.
\begin{equation*}
y=exp\left[-\int P(x)\,dx +C\right] = exp\left[-\int P(x)\,dx\right]e^{C}
\end{equation*}
or,
\begin{equation*}
y=A exp\left[-\int P\,dx\right]
\end{equation*}
where \(A=e^{C}\text{.}\)
\begin{equation*}
\text{or,} \quad ye^{\int P\,dx}=A
\end{equation*}
Let,
\begin{equation*}
I = \int P(x)\,dx \quad \text{so that}\quad \frac{\,dI}{\,dx} =P(x)
\end{equation*}
\begin{equation}
\therefore \quad ye^{I}=A\tag{4.1.7}
\end{equation}
Differentiate eqn. (4.1.7)w.r.t. ’x’, we get -
\begin{equation*}
\frac{\,d}{\,dx}\left(ye^{I}\right)=0
\end{equation*}
or,
\begin{equation*}
\frac{\,dy}{\,dx}e^{I}+ye^{I}\frac{\,dI}{\,dx}=0
\end{equation*}
or,
\begin{equation*}
\frac{dy}{dx}e^{I}+ye^{I}P=0
\end{equation*}
or,
\begin{equation}
e^{I}\left(\frac{dy}{dx}+yP\right)=0\tag{4.1.8}
\end{equation}
Now multiplying eqn. (4.1.6) by \(e^{I} \text{,}\) we get -
\begin{equation}
\left(\frac{\,dy}{\,dx}+Py\right)e^{I}=Qe^{I} \tag{4.1.9}
\end{equation}
which by using eqn. (4.1.9), becomes
\begin{equation}
\frac{\,d}{\,dx}\left(ye^{I}\right)=Qe^{I}\tag{4.1.10}
\end{equation}
Integrating, we get -
\begin{equation*}
ye^{I}=\int Qe^{I}\,dx+C_{1}
\end{equation*}
where \(C_{1}\) is an arbitrary constant.
\begin{equation}
\text{or,} \quad ye^{\int P\,dx}=\int\left(Qe^{\int P\,dx}\right)\,dx+C_{1}\tag{4.1.11}
\end{equation}
This is the required solution and the factor \(e^{\int P\,dx}\) is known as integration factor (I.F.). Therefore the solution of first order linear differential equation can be easily found by the formula
\begin{equation*}
y[I.F.] = \int Q[I.F.]\,dx + \quad\text{constant}.
\end{equation*}
Equation (4.1.11) may be written as
\begin{equation}
y=C_{1}e^{-\int P\,dx}+e^{-\int P\,dx}\int\left(Qe^{\int P\,dx}\right)\,dx\tag{4.1.12}
\end{equation}
i.e. the general solution of differential equation (4.1.6) consists of two parts.
\begin{equation}
y=cu+v \quad \text{(say)}\tag{4.1.13}
\end{equation}
where \(u= e^{-\int P\,dx }\) and \(v=e^{-\int P\,dx}\int\left(Qe^{\int P\,dx}\right)\,dx\) also \(C_{1}=c \text{.}\)
Now differentiating \(u= e^{-\int P\,dx }\) with respect to ’x’, we have
\begin{equation*}
\frac{\,du}{\,dx} =-P e^{-\int P\,dx}=-Pu
\end{equation*}
or,
\begin{equation*}
\frac{\,du}{\,dx} +Pu =0
\end{equation*}
or,
\begin{equation*}
\frac{\,d(cu)}{\,dx} +P(cu) =0
\end{equation*}
which shows that \(y=cu\) is the solution of \(\frac{\,du}{\,dx} +Pu =0\text{,}\) again, differentiating
\begin{equation*}
v=e^{-\int P\,dx}\int\left(Qe^{\int P\,dx}\right)\,dx
\end{equation*}
w.r.t. ’x’, we have -
\begin{equation*}
\frac{\,dv}{\,dx}=-Pe^{-\int P\,dx}\left(Qe^{\int P\,dx}\right)\,dx + e^{-\int P\,dx}Qe^{\int P\,dx}
\end{equation*}
or,
\begin{equation*}
\frac{\,dv}{\,dx} =-Pv+Q
\end{equation*}
\begin{equation*}
\text{or,}\quad \frac{\,dv}{\,dx} +Pv=Q
\end{equation*}
which shows that \(y=v\) is the solution of \(\frac{\,dy}{\,dx} +Py=Q\text{.}\) Hence, the first part of the solution of differential equation is the solution of its homogeneous equation and \(cu \) is known as Complementry Function. Second part is free from any arbitrary constant and \(v\) is known as Particular Integral. \(\therefore\) Complete Solution = Complementry Function + Particular Integral. That is,
\begin{equation*}
y = C.F. + P.I.
\end{equation*}
