The general solution of a differential equation of the type $\frac{d x}{d y}+\mathrm{P}_{1} x=\mathrm{Q}_{1}$ is
A. $y e^{\int \mathrm{P}_{1} d y}=\int\left(\mathrm{Q}_{1} e^{\int \mathrm{P}_{1} d y}\right) d y+\mathrm{C}$
B. $y \cdot e^{\int \mathrm{P}_{1} d x}=\int\left(\mathrm{Q}_{1} e^{\int \mathrm{P}_{1} d x}\right) d x+\mathrm{C}$
C. $x e^{\int \mathrm{P}_{1} d y}=\int\left(\mathrm{Q}_{1} e^{\int \mathrm{P}_{1} d y}\right) d y+\mathrm{C}$
D. $x e^{\int p_{1} d x}=\int\left(Q_{1} e^{\int p_{1} d x}\right) d x+C$
The integrating factor of the given differential equation $\frac{d x}{d y}+\mathrm{P}_{1} x=\mathrm{Q}_{1}$ is $e^{\int p d y}$.
The general solution of the differential equation is given by,
$x(\mathrm{I} . \mathrm{F} .)=\int(\mathrm{Q} \times \mathrm{I} . \mathrm{F} .) d y+\mathrm{C}$
$\Rightarrow x \cdot e^{\int P d y}=\int\left(Q_{1} e^{\int P_{p d y}}\right) d y+\mathrm{C}$
Hence, the correct answer is C.
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