The general solution of a differential equati | vedclass

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(A) The given differential equation is of the form $\frac{dx}{dy} + P_{1}x = Q_{1}$,where $P_{1}$ and $Q_{1}$ are functions of $y$ only.The integratin

Vedclass · Accepted Answer

$x \cdot e^{\int P_{1} dy} = \int (Q_{1} \cdot e^{\int P_{1} dy}) dy + C$

Vedclass · Answer

(A) The given differential equation is of the form $\frac{dx}{dy} + P_{1}x = Q_{1}$,where $P_{1}$ and $Q_{1}$ are functions of $y$ only.The integrating factor ($I$.$F$.) for this differential equation is given by:$I.F. = e^{\int P_{1} dy}$The general solution of the differential equation is given by:$x \cdot (I.F.) = \int (Q_{1} \cdot I.F.) dy + C$Substituting the value of the integrating factor:$x \cdot e^{\int P_{1} dy} = \int (Q_{1} \cdot e^{\int P_{1} dy}) dy + C$Comparing this with the given options,the correct option is $A$.

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

The general solution of a differential equation of the type $\frac{dx}{dy} + P_{1}x = Q_{1}$ is

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