Thank you for reporting, we will resolve it shortly
Q.
The general solution of the first order linear differential equation of the type $\frac{d y}{d x}+P y=Q$ is
Differential Equations
Solution:
To solve the first order linear differential equation of the type
$\frac{d y}{d x}+P y=Q ....$(i)
On multiplying both sides of the equation by a function of $x$ say $g(x)$ to get
$g(x) \frac{d y}{d x}+P \cdot(g(x)) y=Q \cdot g(x)$...(ii)
Choose $g(x)$ in such a way that $RHS$ becomes a derivative of $y \cdot g(x)$.
i.e., $ g(x) \frac{d y}{d x}+P \cdot g(x) y =\frac{d}{d x}[y \cdot g(x)]$
$\Rightarrow g(x) \frac{d y}{d x}+P \cdot g(x) y =g(x) \frac{d y}{d x}+y g^{\prime}(x)$
$\Rightarrow P(x) =g^{\prime}(x)$
or $ P =\frac{g^{\prime}(x)}{g(x)}$
Integrating both sides w.r.t. $x$, we get
$ \int P d x =\int \frac{g^{\prime}(x)}{g(x)} d x$
$\Rightarrow \int P \cdot d x =\log (g(x))$
$\Rightarrow g(x) =e^{\int P d x}$
On multiplying the Eq. (i) by $g(x)=e^{\int P d x}$, the LHS becomes the derivative of some function of $x$ and $y$. This function $g(x)=e^{\int P d x}$ is called Integrating Factor (IF) of the given differential equation.
On substituting the value of $g(x)$ in Eq. (ii), we get
$e^{\int P d x} \frac{d y}{d x}+P e^{\int P d x} y=Q \cdot e^{\int P d x}$
$\Rightarrow \frac{d}{d x}\left(y e^{\int P d x}\right)=Q e^{\int P d x}$
On integrating both sides w.r.t. $x$, we get
$y \cdot e^{\int P d x} =\int\left(Q \cdot e^{\int P d x}\right) d x$
$\Rightarrow y =e^{-\int P d x} \cdot \int\left(Q e e^{\int P d x}\right) d x+C$
which is the general solution of the differential equation.