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The integrating factor of the linear directional equation $\frac{d y}{d x}+P(x) y=Q(x)$ is a solution of the differential equation
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Verified Answer
The correct answer is:
$\frac{d y}{d x}-P(x) y=0$
$\frac{d y}{d x}+P(x) y=Q(x)$
$\begin{aligned} & \text { Integrating }=e^{\int P(x) d x} \\ & \text { Factor }\end{aligned}$
(A) If $y=e^{\int p(x) d x}$
checking:
$\frac{d y}{d x}-P(x) y=0$
$\frac{d y}{d x}=e^{\int {p}(x) d x}$ $P(x)=y P(x)$
$\Rightarrow P(x) y-y P(x)=0$
$\Rightarrow$ Hence found
$\Rightarrow$ If of $\quad \frac{d y}{d x}+P(x) y=Q(x)$ i.e., $e^{\int p(x) d x}$ is the solution of (A)
ie;,$\frac{d y}{d x}-P(x) y=0$
$\begin{aligned} & \text { Integrating }=e^{\int P(x) d x} \\ & \text { Factor }\end{aligned}$
(A) If $y=e^{\int p(x) d x}$
checking:
$\frac{d y}{d x}-P(x) y=0$
$\frac{d y}{d x}=e^{\int {p}(x) d x}$ $P(x)=y P(x)$
$\Rightarrow P(x) y-y P(x)=0$
$\Rightarrow$ Hence found
$\Rightarrow$ If of $\quad \frac{d y}{d x}+P(x) y=Q(x)$ i.e., $e^{\int p(x) d x}$ is the solution of (A)
ie;,$\frac{d y}{d x}-P(x) y=0$
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