Match the differential equations in List I to | vedclass

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Question

(A) For a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$, the integrating factor $(IF)$ is given by $e^{\int P(x) dx}$.$(P)$

Vedclass · Accepted Answer

$P-5, Q-1, R-2, S-3$

Vedclass · Answer

(A) For a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$, the integrating factor $(IF)$ is given by $e^{\int P(x) dx}$.$(P)$ $(x^3+1)\frac{dy}{dx}+x^2y=3x^2 \Rightarrow \frac{dy}{dx} + \frac{x^2}{x^3+1}y = \frac{3x^2}{x^3+1}$.$IF = e^{\int \frac{x^2}{x^3+1} dx} = e^{\frac{1}{3}\ln(x^3+1)} = (x^3+1)^{1/3}$. So, $P-5$.$(Q)$ $x^2\frac{dy}{dx}+3xy=x^6 \Rightarrow \frac{dy}{dx} + \frac{3}{x}y = x^4$.$IF = e^{\int \frac{3}{x} dx} = e^{3\ln x} = x^3$. So, $Q-1$.$(R)$ $(x^3+1)^2\frac{dy}{dx}+6x^2(x^3+1)y=x^2 \Rightarrow \frac{dy}{dx} + \frac{6x^2}{x^3+1}y = \frac{x^2}{(x^3+1)^2}$.$IF = e^{\int \frac{6x^2}{x^3+1} dx} = e^{2\ln(x^3+1)} = (x^3+1)^2$. So, $R-2$.$(S)$ $(x^2+1)\frac{dy}{dx}+4xy=\ln x \Rightarrow \frac{dy}{dx} + \frac{4x}{x^2+1}y = \frac{\ln x}{x^2+1}$.$IF = e^{\int \frac{4x}{x^2+1} dx} = e^{2\ln(x^2+1)} = (x^2+1)^2$. So, $S-3$.Thus, the correct match is $P-5, Q-1, R-2, S-3$.

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

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