Question

Determine whether this differential equations is an exact differential equation. and if exact give the solution ((X^2)/2)log(y)+ ((x^3)/6y) dy/dx =0 y(1)=1

Answer 1

\[(\frac{ x^{2} }{ 2 }*\ln(y))+(\frac{ x^{3} }{ 6y })*\frac{ dy }{ dx }=0\]

Correct?

Answer 2

Im currently learning it so if u could show how u got to ur answer, which is probably right. Much appreciated.

Answer 3

No, I was asking if I got the question you typed up correct. I wanted to make sure I was reading it correctly xD Im pretty new to it myself, but I think Ic an help.

Answer 4

then yes im sorry im pretty tired
and y(1)= 1

Answer 5

Alright, So ill just multiply everything by dx to start, that way I can visually confirm which part is my M function and which is my N. \[\frac{ x^{2}lny }{ 2}dx+\frac{ x^{3} }{ 6y }dy \]So, the part of the equation that takes on dx is our "M" function and the part that takes on dy is our "N" function. If the DE is exact, then the partial derivative of M with respect to y will be the same as the partial derivative of N with respect to x. I just remember which variable is what in these equations because you'll always take the partial derivative with respect to the variable of the derivative not attached to the term. Basically, I mean that dx means partial to y, dy means partial to x. And you always integrate with respect to the same variable. dx means integrate with respect to x, dy means integrate with respect to y. So thats just how I memorize the order of these things. Okay, now to check for exactness. Partial of M with respect to y comes out to be:
\[\frac{ x^{2} }{ 2y } \]And the partial of N with respect to x is:
\[\frac{ x^{2} }{ 2y }\]And if you want, I can show you the parital derivative stuff. Do you know how to get the partials?

Answer 6

Ok i understand what you're saying but if you dont mind showing the partials

Answer 7

Yeah, sure. Well, when I say take the partial with respect to y, that means I pretend all the x's in the equation are constants. So if I had the partial derivative with respect to y of (x+y), the x would disappear just like a constant would. Of course having all the random letters can be confusing, so I like to block out all the x values, or y-values if its respect to x. So for the first part, x^2lny/2, because this is the dx term, I take the derivative with respect to y. As to not confuse myself, Im just going to pretend the x^2/2 part isnt even there. So the derivative of lny is just 1/y. Because none of the numbers or x's were by themselves, none of them disappeared, they just get tacked on to the 1/y, meaning that partial derivative is x^2/2y

Now the other one, because it has a dy tacked on to it, is taken with respect to x. This means I want to ignore all the y variables in the term. If I ignore everything but x, I have x^3, which the derivative is 3x^2 of course. Now if I bring back the part of the term I ignored, I have 3x^2/6y, which simplifies to x^2/6y. So both of those partial derivatives gave the same result, meaning this is an exact DE.

Answer 8

Wow thank you so much been of much help

Answer 9

Do you know how to take it from here, or do you wanna work it out more?

Answer 10

There is more? I thought i was confident for no reason. If you could please. Work out all the way.

Answer 11

Yeah, all we did was proveit is exact, but we did nothign to solve it. If it is exact, we can use a specific method. If it were not exact, we would have to see if there was some other way. So now that weve shown its exact, these are kind of the steps. Ill number them out then work them out:
1. Integrate "M" with respect to x. The constant of "C" is actually not just a constant, but an entire function of y. So after we integrate, we put g(y) instead of C.
2. Take the derivative of what you just integrated with respect to y. This will give you some function + g'(y) (derivative of the constant).
3. Take what you have now and set all of it equal to original "N" function and solve for g'(y).
4. Integrate g'(y) and what you found g'(y) to be equal to. This will give you the value of g(y) which we can plug back into the result of step 1. All of this ends up being equal to "c". A lot of steps, but youll see.

Answer 12

So first step is integrate "M" with respect to x. So we have:
\[\int\limits_{}^{}\frac{ x ^{2}lny }{ 2 }dx = \frac{ lny }{ 2 }\int\limits_{}^{}x^{2}dx\]
Since im taking the derivative with respect to x, all that lny/2 stuff is just a constant. So when I integrate and multiply the lny/2 back in, I have:

\[\frac{ x^{3}lny }{ 6 }+g(y)\]And remember that this time the constant is g(y) and not just C. Just to explain that portion, we integrated with respect to x, meaning that y's were just constant. But the y's didnt come from nowhere, there had to be some derivative fromw hich theycame from. So g(y) is just our dummy variable that represents whatever function the y's came from.

So next step is to now differentiate with respect to y. SO pretending the x^3/6 portion is just a constant, we get:

\[\frac{ x^{3} }{ 6y }+g'(y)\]

Now the next step is to set this result equal to the original function "N" and solve for g'(y). So doing that we have:

\[\frac{ x^{3} }{ 6y }+g'(y) = \frac{ x^{3} }{ 6y }\implies g'(y) = 0\]
Well, that made life easy. If g'(y) is 0, then we cant integrate it, thatll also be 0. So there is no g(y) and what we integrated earlier is our answer. So that means wehave simply:

\[\frac{ x^{3}lny }{ 6y }+g(y) \implies \frac{ x^{3}lny }{ 6y }+0 = C\]Now we were given an initial condition, so that means we can find what our C is. y(1) = 1 is the condition. THis basically is saying "when x is 1, y is 1", So all we dois plug in 1 for x and for y to get:

\[\frac{ 1^{3}\ln(1) }{ 6(1) }= C \implies C = 0\]So even more simplicity, no g(y), no C, so our final answer is simply

\[\frac{ x^{3}lny }{ 6y }\]