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(C) Given the differential equation: $x^2 f^{\prime}(x) - 2 x f(x) = 3$.Divide both sides by $x^4$ (where $x 
eq 0$):$\frac{x^2 f^{\prime}(x) - 2 x f

Vedclass · Accepted Answer

$39$

Vedclass · Answer

(C) Given the differential equation: $x^2 f^{\prime}(x) - 2 x f(x) = 3$.Divide both sides by $x^4$ (where $x 
eq 0$):$\frac{x^2 f^{\prime}(x) - 2 x f(x)}{x^4} = \frac{3}{x^4}$This simplifies to the derivative of a quotient:$\frac{d}{dx} \left( \frac{f(x)}{x^2} ight) = 3 x^{-4}$Integrating both sides with respect to $x$:$\frac{f(x)}{x^2} = \int 3 x^{-4} dx = 3 \left( \frac{x^{-3}}{-3} ight) + C = -\frac{1}{x^3} + C$Multiplying by $x^2$:$f(x) = -\frac{1}{x} + C x^2$Given $f(1) = 4$,substitute $x=1$:$4 = -\frac{1}{1} + C(1)^2 \Rightarrow 4 = -1 + C \Rightarrow C = 5$.Thus,$f(x) = 5x^2 - \frac{1}{x}$.Now,calculate $2f(2)$:$f(2) = 5(2)^2 - \frac{1}{2} = 20 - 0.5 = 19.5$.$2f(2) = 2 	imes 19.5 = 39$.

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

Let $f :(0, \infty) \rightarrow R$ be a function which is differentiable at all points of its domain and satisfies the condition $x^2 f^{\prime}(x)=2 x f(x)+3$,with $f(1)=4$. Then $2 f(2)$ is equal to:

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