The general solution of the differential equa | vedclass

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(A) Given the differential equation: $(1-x^{2}) \frac{dy}{dx} + 2xy = x(1-x^{2})^{\frac{1}{2}}$Divide by $(1-x^{2})$ to get the standard form $\frac{d

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$y = \sqrt{1-x^{2}} + c(1-x^{2})$

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(A) Given the differential equation: $(1-x^{2}) \frac{dy}{dx} + 2xy = x(1-x^{2})^{\frac{1}{2}}$Divide by $(1-x^{2})$ to get the standard form $\frac{dy}{dx} + Py = Q$:$\frac{dy}{dx} + \frac{2x}{1-x^{2}}y = \frac{x}{\sqrt{1-x^{2}}}$Here,$P = \frac{2x}{1-x^{2}}$ and $Q = \frac{x}{\sqrt{1-x^{2}}}$.Integrating Factor $(I.F.) = e^{\int P dx} = e^{\int \frac{2x}{1-x^{2}} dx} = e^{-\ln(1-x^{2})} = e^{\ln(\frac{1}{1-x^{2}})} = \frac{1}{1-x^{2}}$.The general solution is $y 	imes (I.F.) = \int Q 	imes (I.F.) dx + c$.$y 	imes \frac{1}{1-x^{2}} = \int \frac{x}{\sqrt{1-x^{2}}} 	imes \frac{1}{1-x^{2}} dx + c$$y 	imes \frac{1}{1-x^{2}} = \int x(1-x^{2})^{-\frac{3}{2}} dx + c$Let $u = 1-x^{2}$,then $du = -2x dx$,so $x dx = -\frac{1}{2} du$.$y 	imes \frac{1}{1-x^{2}} = -\frac{1}{2} \int u^{-\frac{3}{2}} du + c = -\frac{1}{2} \left( \frac{u^{-\frac{1}{2}}}{-\frac{1}{2}} ight) + c = u^{-\frac{1}{2}} + c = \frac{1}{\sqrt{1-x^{2}}} + c$.Multiplying both sides by $(1-x^{2})$:$y = \frac{1-x^{2}}{\sqrt{1-x^{2}}} + c(1-x^{2}) = \sqrt{1-x^{2}} + c(1-x^{2})$.

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 the differential equation $(1-x^{2}) \frac{dy}{dx} + 2xy = x(1-x^{2})^{\frac{1}{2}}$ is

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