The general solution of the differential equa | vedclass

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(B) Step $1$: Rewrite the equation as $\frac{dx}{dy} + x \cos y = e^{\sin y} \cos y$.Step $2$: This is a linear differential equation of the form $\fr

Vedclass · Accepted Answer

$xe^{\sin y} = \frac{e^{2 \sin y}}{2} + c$

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(B) Step $1$: Rewrite the equation as $\frac{dx}{dy} + x \cos y = e^{\sin y} \cos y$.Step $2$: This is a linear differential equation of the form $\frac{dx}{dy} + Px = Q$, where $P = \cos y$ and $Q = e^{\sin y} \cos y$.Step $3$: Find the integrating factor $IF = e^{\int P dy} = e^{\int \cos y dy} = e^{\sin y}$.Step $4$: The solution is $x(IF) = \int Q(IF) dy + c$.Step $5$: $x e^{\sin y} = \int (e^{\sin y} \cos y) (e^{\sin y}) dy + c = \int e^{2 \sin y} \cos y dy + c$.Step $6$: Let $u = \sin y$, then $du = \cos y dy$. The integral becomes $\int e^{2u} du = \frac{e^{2u}}{2} = \frac{e^{2 \sin y}}{2}$.Step $7$: Thus, $x e^{\sin y} = \frac{e^{2 \sin y}}{2} + c$.

The general solution of the differential equation $\sec y + (x - e^{\sin y}) \frac{dy}{dx} = 0$ is...

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