By multiplying with e^{int P dx} on both side | vedclass

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(C) The given linear differential equation is $\frac{dy}{dx} + P(x)y = Q(x)$.Multiplying both sides by the integrating factor $I.F. = e^{\int P dx}$,w

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$e^{\int P dx}$

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(C) The given linear differential equation is $\frac{dy}{dx} + P(x)y = Q(x)$.Multiplying both sides by the integrating factor $I.F. = e^{\int P dx}$,we get:$e^{\int P dx} \frac{dy}{dx} + y P(x) e^{\int P dx} = Q(x) e^{\int P dx}$.We know that the derivative of the product of $y$ and the integrating factor is:$\frac{d}{dx}(y e^{\int P dx}) = e^{\int P dx} \frac{dy}{dx} + y \frac{d}{dx}(e^{\int P dx})$.Since $\frac{d}{dx}(e^{\int P dx}) = e^{\int P dx} P(x)$,the expression becomes:$\frac{d}{dx}(y e^{\int P dx}) = e^{\int P dx} \frac{dy}{dx} + y e^{\int P dx} P(x)$.Comparing this with the given form $\frac{d}{dx}(y f(x))$,we identify that $f(x) = e^{\int P dx}$.

List $I$ (Differential Equation)	List $II$ (Integrating Factor)
$(P)$ $(x^3+1)\frac{dy}{dx}+x^2y=3x^2$	$(1)$ $x^3$
$(Q)$ $x^2\frac{dy}{dx}+3xy=x^6$	$(2)$ $(x^3+1)^2$
$(R)$ $(x^3+1)^2\frac{dy}{dx}+6x^2(x^3+1)y=x^2$	$(3)$ $(x^2+1)^2$
$(S)$ $(x^2+1)\frac{dy}{dx}+4xy=\ln x$	$(4)$ $x^2+1$
	$(5)$ $(x^3+1)^{1/3}$
	$(6)$ $(x^3+1)^{1/2}$

By multiplying with $e^{\int P dx}$ on both sides of the equation $\frac{dy}{dx} + P(x)y = Q(x)$,the left side of the equation takes the form $\frac{d}{dx}(y f(x))$,then $f(x) =$

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