Finding the differential equation.

I didn't know equations had unique differential equations.

Do you mean, what is the solution to this differential equation? It's separable.
 
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
sin(y)+xcos(y)y=cecx\displaystyle \sin(y)+ xcos(y)y'= ce^{cx}, a differential equation satisfied that expression.
 
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
sin(y)+xcos(y)y=cecx\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.
 
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
 
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 xsin(y)=ecx\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 dF(x,y)dx=xsin(y)ecx\displaystyle \frac{dF(x,y)}{dx}=x sin(y)- e^{cx}.
 
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 xsin(y)=ecx\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 dF(x,y)dx=xsin(y)ecx\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.
 
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 "sin(y)+xcos(y)dydx=cecx\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 xsin(y)=ecy\displaystyle xsin(y)= e^{cy} to write sin(y)+xcos(y)dydx=cxsin(y)\displaystyle sin(y)+ xcos(y)\frac{dy}{dx}= c x sin(y). Finally from the original equation, ln(xsin(y))=cy\displaystyle ln(x sin(y))= cy so c=ln(xsin(y)y\displaystyle c= \frac{ln(x sin(y)}{y} and then sin(y)+xcos(y)dydx=ln(xsin(y))xsin(y)y\displaystyle sin(y)+ xcos(y)\frac{dy}{dx}= \frac{ ln(x sin(y)) x sin(y)}{y}.
 
Top