The general solution of frac{dy}{dx} + y tan | vedclass

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(D) The given differential equation is $\frac{dy}{dx} + y 	an x = 2x + x^2 	an x$. This is a linear differential equation of the form $\frac{dy}{dx}

Vedclass · Accepted Answer

$y = x^2 + c \cos x$

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(D) The given differential equation is $\frac{dy}{dx} + y 	an x = 2x + x^2 	an x$. This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = 	an x$ and $Q = 2x + x^2 	an x$. First,we find the Integrating Factor ($I$.$F$.): $I.F. = e^{\int P dx} = e^{\int 	an x dx} = e^{\ln |\sec x|} = \sec x$. The general solution is given by: $y \cdot (I.F.) = \int Q \cdot (I.F.) dx + c$ $y \sec x = \int (2x + x^2 	an x) \sec x dx + c$ $y \sec x = \int 2x \sec x dx + \int x^2 	an x \sec x dx + c$ Using integration by parts for the second integral $\int x^2 (	an x \sec x) dx$: Let $u = x^2$ and $dv = \sec x 	an x dx$. Then $du = 2x dx$ and $v = \sec x$. $\int x^2 \sec x 	an x dx = x^2 \sec x - \int 2x \sec x dx$. Substituting this back into the equation: $y \sec x = \int 2x \sec x dx + (x^2 \sec x - \int 2x \sec x dx) + c$ $y \sec x = x^2 \sec x + c$ Multiplying both sides by $\cos x$: $y = x^2 + c \cos x$.

The general solution of $\frac{dy}{dx} + y \tan x = 2x + x^2 \tan x$ is:

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