#### jerelova24

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What is the DE of xsiny=e^(cx)

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What is the DE of xsiny=e^(cx)

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We can, of course, differentiate the given expression with respect to x to get

\(\displaystyle \sin(y)+ xcos(y)y'= ce^{cx}\), a differential equation satisfied that expression.

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Well, sorry for abuse. I was just going with the zeroth derivative as it was presented.

We can, of course, differentiate the given expression with respect to x to get

\(\displaystyle \sin(y)+ xcos(y)y'= ce^{cx}\), a differential equation satisfied that expression.

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Should be sin(y)+xcos(y)y′=cexsin(y)+xcos(y)y′=ce^{x}

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Yes, of course. Thank you. (It's been corrected).

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I'm still waiting for the answer to the question: What do you mean by this? There are a large number (in fact, an infinite number) of differential equations you can write where this is the solution.What is the DE of xsiny=e^(cx)

Is there more to this problem?

-Dan

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But what do you mean by this? The "zeroth derivative" of a function is just that function. In your original post you had the equation \(\displaystyle x sin(y)= e^{cx}\), an equality ofI was just going with the zeroth derivative as it was presented.

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Over-thinking. It was an ill-formed question and I provided an similarly-motivated response. No rigor intended.But what do you mean by this? The "zeroth derivative" of a function is just that function. In your original post you had the equation \(\displaystyle x sin(y)= e^{cx}\), an equality oftwofunctions. I still don't understand what you want. Are you asking for an "anti-derivative, a function, f(x,y)= Constant, such that \(\displaystyle \frac{dF(x,y)}{dx}=x sin(y)- e^{cx}\).

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We can then use the original equation \(\displaystyle xsin(y)= e^{cy}\) to write \(\displaystyle sin(y)+ xcos(y)\frac{dy}{dx}= c x sin(y)\). Finally from the original equation, \(\displaystyle ln(x sin(y))= cy\) so \(\displaystyle c= \frac{ln(x sin(y)}{y}\) and then \(\displaystyle sin(y)+ xcos(y)\frac{dy}{dx}= \frac{ ln(x sin(y)) x sin(y)}{y}\).