The solution of the differential equation (1 | vedclass

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Question

(C) Given the differential equation: $(1 + x^2)\frac{dy}{dx} = x$Separate the variables:$dy = \frac{x}{1 + x^2} dx$Integrate both sides:$\int dy = \in

Vedclass · Accepted Answer

$y = \frac{1}{2}\log_e(1 + x^2) + c$

Vedclass · Answer

(C) Given the differential equation: $(1 + x^2)\frac{dy}{dx} = x$Separate the variables:$dy = \frac{x}{1 + x^2} dx$Integrate both sides:$\int dy = \int \frac{x}{1 + x^2} dx + c$Let $u = 1 + x^2$,then $du = 2x dx$,which implies $x dx = \frac{1}{2} du$.Substituting this into the integral:$y = \int \frac{1}{2u} du + c$$y = \frac{1}{2} \log_e|u| + c$$y = \frac{1}{2} \log_e(1 + x^2) + c$Thus,the correct option is $C$.

The solution of the differential equation $(1 + x^2)\frac{dy}{dx} = x$ is

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