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Question

(B) Given the differential equation: $\frac{dy}{dx} = e^x + \cos x + x + 	an x$To find the solution,we integrate both sides with respect to $x$:$y =

Vedclass · Accepted Answer

$y = e^x + \sin x + \frac{x^2}{2} + \log \sec x + c$

Vedclass · Answer

(B) Given the differential equation: $\frac{dy}{dx} = e^x + \cos x + x + 	an x$To find the solution,we integrate both sides with respect to $x$:$y = \int (e^x + \cos x + x + 	an x) dx$Using the standard integrals:$\int e^x dx = e^x$$\int \cos x dx = \sin x$$\int x dx = \frac{x^2}{2}$$\int 	an x dx = \log |\sec x|$Combining these,we get:$y = e^x + \sin x + \frac{x^2}{2} + \log |\sec x| + c$Thus,the correct option is $B$.

The solution of the differential equation $\frac{dy}{dx} = e^x + \cos x + x + \tan x$ is

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