The solution of cos y frac{dy}{dx} = e^{x+sin | vedclass

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(D) Given the differential equation: $\cos y \frac{dy}{dx} = e^x e^{\sin y} + x^2 e^{\sin y}$.Divide both sides by $e^{\sin y}$: $e^{-\sin y} \cos y \

Vedclass · Accepted Answer

$e^x + \frac{1}{3} x^3$

Vedclass · Answer

(D) Given the differential equation: $\cos y \frac{dy}{dx} = e^x e^{\sin y} + x^2 e^{\sin y}$.Divide both sides by $e^{\sin y}$: $e^{-\sin y} \cos y \frac{dy}{dx} = e^x + x^2$.Let $u = \sin y$, then $\frac{du}{dx} = \cos y \frac{dy}{dx}$.The equation becomes $e^{-u} \frac{du}{dx} = e^x + x^2$.Integrating both sides with respect to $x$: $\int e^{-u} du = \int (e^x + x^2) dx$.$-e^{-u} = e^x + \frac{x^3}{3} + C_1$.Substitute $u = \sin y$ back: $-e^{-\sin y} = e^x + \frac{x^3}{3} + C_1$.Rearranging to the form $f(x) + e^{-\sin y} = C$: $e^x + \frac{x^3}{3} + e^{-\sin y} = C$.Comparing this with $f(x) + e^{-\sin y} = C$, we get $f(x) = e^x + \frac{x^3}{3}$.

The solution of $\cos y \frac{dy}{dx} = e^{x+\sin y} + x^2 e^{\sin y}$ is $f(x) + e^{-\sin y} = C$ ($C$ is an arbitrary real constant), where $f(x)$ is equal to:

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