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(A) Given the differential equation: $\frac{dy}{dx} = \frac{1}{xy(x^2 \sin y^2 + 1)}$.Rearranging,we get: $\frac{dx}{dy} = xy(x^2 \sin y^2 + 1) = x^3

Vedclass · Accepted Answer

$e^{y^2} \left( \frac{1}{x^2} - \frac{\cos y^2}{2} + \frac{\sin y^2}{2} \right) = C$

Vedclass · Answer

(A) Given the differential equation: $\frac{dy}{dx} = \frac{1}{xy(x^2 \sin y^2 + 1)}$.Rearranging,we get: $\frac{dx}{dy} = xy(x^2 \sin y^2 + 1) = x^3 y \sin y^2 + xy$.Divide by $x^3$: $x^{-3} \frac{dx}{dy} - yx^{-1} = y \sin y^2$.Let $u = x^{-2}$,then $\frac{du}{dy} = -2x^{-3} \frac{dx}{dy}$,so $x^{-3} \frac{dx}{dy} = -\frac{1}{2} \frac{du}{dy}$.Substituting into the equation: $-\frac{1}{2} \frac{du}{dy} - yu = y \sin y^2 \Rightarrow \frac{du}{dy} + 2yu = -2y \sin y^2$.This is a linear differential equation in $u$ with integrating factor $IF = e^{\int 2y dy} = e^{y^2}$.The solution is $u e^{y^2} = \int (-2y \sin y^2) e^{y^2} dy$.Let $t = y^2$,then $dt = 2y dy$. The integral becomes $\int -\sin t e^t dt$.Using integration by parts: $\int e^t \sin t dt = \frac{e^t}{2} (\sin t - \cos t)$.So,$u e^{y^2} = -\frac{e^{y^2}}{2} (\sin y^2 - \cos y^2) + C$.Substituting $u = x^{-2}$: $\frac{e^{y^2}}{x^2} = \frac{e^{y^2}}{2} (\cos y^2 - \sin y^2) + C$.Rearranging gives: $e^{y^2} \left( \frac{1}{x^2} - \frac{\cos y^2}{2} + \frac{\sin y^2}{2} \right) = C$.

Find the solution of the differential equation $\frac{dy}{dx} = \frac{1}{xy(x^2 \sin y^2 + 1)}$,where $C$ is the integral constant.

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