Question

solve :(y cos x+1)DX+(sin x)dy=0​

Answer 1

$\underline{\textbf{Given:}}$

$\mathsf{(y\;cosx+1)dx+sinx\;dy=0}$

$\underline{\textbf{To find:}}$

$\textsf{solution of the given differential equation}$

$\underline{\textbf{Solution:}}$

$\underline{\textbf{Product rule of differentiation:}}$

$\boxed{\mathsf{\dfrac{d(uv)}{dx}=u\;\dfrac{dv}{dx}+v\;\dfrac{du}{dx}}}$

$\mathsf{Consider,}$

$\mathsf{(y\;cosx+1)dx+sinx\;dy=0}$

$\textsf{This can be written as}$

$\mathsf{y\;cosx+1+sinx\;\dfrac{dy}{dx}=0}$

$\mathsf{y\;cosx+sinx\;\dfrac{dy}{dx}=-1}$

$\textsf{Using Product rule, we get}$

$\mathsf{\dfrac{d(y\;sinx)}{dx}=-1}$

$\mathsf{d(y\;sinx)=-dx}$

$\textsf{Integrating,}$

$\mathsf{\int\;d(y\;sinx)=-\int\;dx}$

$\mathsf{y\;sinx=-x+C}$

$\implies\boxed{\mathsf{x+(y\;sinx)=C}}$

$\textsf{which is the required solution}$

$\underline{\textbf{Find more:}}$

Solve x(x-y)dy+y^2dx=0​

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For DE, (y cos x)dx + (x cos y) dy =0, test for exactness requires that
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