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  • The general solution of ex cosy dx – ex siny dy = 0 is: A. ex cosy = k B. ex siny = k C. ex = k cosy D. ex = k siny

The general solution of e^x \cos y dx - e^x \sin y dy = 0 is:
A. e^x \cos y = k
B. e^x \sin y = k
C. e^x = k \cos y
D. e^x = k \sin y

Answers (1)

\\e^{x} \cos y d x-e^{x} \sin y d y=0$ \\$e^{x} \cos y d x=e^{x} \sin y d y$ \\$\Rightarrow d x=\frac{\sin y}{\cos y} d y
Integrate
\Rightarrow \int \mathrm{dx}=\int \frac{\sin y}{\cos y} \mathrm{~d} y
substitute cosy =t hence
\frac{\mathrm{dt}}{\mathrm{dy}}=-\sin \mathrm{y}
Which means sinydy=-dt
\\\quad \Rightarrow x=\int \frac{-d t}{t}$ \\$x=-\log t+c$ \\$x+c=\log (\cos y)^{-1}

\\ \Rightarrow x+c=\log \frac{1}{\cos y}$ \\$x+c=\log (\sec y)$ \\$e^{x+c}=\sec y$ \\$\mathrm{e}^{x } \mathrm{e}^{c}=\sec y$ \\$\Rightarrow \mathrm{e}^{\mathrm{x}}=\frac{1}{\mathrm{e}^{\mathrm{c}} \operatorname{cosy}}$ \\$e^{x} \cos y=e^{-c}$ \\$\mathrm{e}^{-\mathrm{c}}$ is constant because e is a constant and \\$\mathrm{c}$ is the integration constant let us it denote as $\mathrm{k}$ hence $\mathrm{e}^{-c}=\mathrm{k}
$$ e^{x} \cos y=k

Option A is correct.

Posted by

infoexpert22

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