Finding the differential equation.

jerelova24

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What is the DE of xsiny=e^(cx)
 

tkhunny

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I didn't know equations had unique differential equations.

Do you mean, what is the solution to this differential equation? It's separable.
 

HallsofIvy

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As given, that is not a differential equation so I don't know what tkhunny means by his last line.

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.
 

tkhunny

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As given, that is not a differential equation so I don't know what tkhunny means by his last line.

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

HallsofIvy

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

topsquark

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

Is there more to this problem?

-Dan
 

HallsofIvy

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I was just going with the zeroth derivative as it was presented.
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 of two functions. I still don't understand what you want. Are you asking for an "anti-derivative, a function, f(x,y)= Constant, such that \(\displaystyle \frac{dF(x,y)}{dx}=x sin(y)- e^{cx}\).
 

tkhunny

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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 of two functions. I still don't understand what you want. Are you asking for an "anti-derivative, a function, f(x,y)= Constant, such that \(\displaystyle \frac{dF(x,y)}{dx}=x sin(y)- e^{cx}\).
Over-thinking. It was an ill-formed question and I provided an similarly-motivated response. No rigor intended.
 

HallsofIvy

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I agre{e that this was an ill formed question. I would assume that the OP meant the simplest DE that has the given function as the general solution. I previously wrote "\(\displaystyle sin(y)+ x cos(y)\frac{dy}{dx}= ce^{cx}\)". Of course, that is not the answer because it still has "c" in it.

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 \(\displaystyle sin(y)+ xcos(y)\frac{dy}{dx}= \frac{ ln(x sin(y)) x sin(y)}{y}\).
 
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