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Mathematics 90 Online
OpenStudy (anonymous):

Have an exact equation problem for you guys: (y^2-2x^2-1)y' - (4xy+x^2+2) = 0 N = (y^2-2x^2-1) => -4x M = -(4xy+x^2+2) => -4x (exact) My answer keeps coming out to: (-2x^2y)-(x^3/3)-(2x)+(y^3/3)-y= c But apparently that's incorrect. I've checked my integration and derivation twice, can someone tell me if they get the same answer or if something is different? Thanks!

OpenStudy (abb0t):

Well if they are exact this is what you need to do, I will walkthrough what you should of done and NOTE down where I made the most common mistakes in and what I remember people did also: If I remember correctly, I believe that you take the integral first of M with respect to x first (or you can also find ∫N dy) to get your psi function: f(x)+h where "h" is like a constant. [NOTE: the “constant” of integration is not really a constant at all, but instead it will be a function of the remaining variable(s), depending on what you integrated to get psi, whether it was M or N]. tHE next step is to differentiate your previous M, and set it equal to N (assuming you chose to integrate M first). [NOTE 2: Don't forget to integrate h(y) also!! Common mistake] you should notice that you should have h'(y) and you should only have ONE single variable. [NOTE 3: If you chose N, then you should have only X's and if you chose M, you should have only Y's! If you have more than one variable, you made a mistake somewhere]. Now, you can find "h" by integrating, now you will get a new constant. However, you can just drop because you will eventually have h-k and that's basically two constants (I think that's the wrong logic, but that's how I see it). Plug everything together for "h" and you can go straight to your implicit solution.

OpenStudy (abb0t):

If you need further clarification, please let me know. I just wanted to lay down everything first to make sure you check your work again because these problems tend to take a bit of time to work out and I'm feeling lazy right now :P

OpenStudy (anonymous):

No, that's the procedure I followed. I'll write my steps out, good job explaining though.

OpenStudy (abb0t):

Ahhh! Lol I haven't taken this course in a while, but I do remember how tedious exact differential equations were :P

OpenStudy (anonymous):

\[\int\limits_{}^{} M dx = \int\limits_{}^{} -4xy-x ^{2}-2dx = -2x ^{2}*y-\frac{ x ^{3} }{ 3 } - 2x + h(y)\] \[\delta y \Psi -2x ^{2}-\frac{ x ^{3} }{3 }-2x+h(y) = -2x ^{2} + h'(y)\] \[\Psi y = -2x ^{2} +h'(y) = y ^{2} -2x^{2} -1 = N\] \[h(y) = y^{2} -1 = \int\limits_{}^{} y^2-1 dy = \frac{ y^{3} }{ 3 } -y +k\] \[-2x^{2}y -\frac{ x^3 }{ 3 } - 2x+\frac{ y^{3} }{3}- y +k = c \] \[OR -2x^{2}y -\frac{ x^3 }{ 3 } - 2x+\frac{ y^{3} }{3}- y= c \]

OpenStudy (abb0t):

That looks correct to me. Mayb the answer looks different because they grouped certain terms together and factored out some stuff. Other than that, it should be an acceptable answer.

OpenStudy (abb0t):

Were you given any initial conditions? If you were, that will definitely give you a different answer.

OpenStudy (anonymous):

Nope. None. At least I can go to bed knowing I'm better then the program I need to enter the answer into. Thanks for checking my work!

OpenStudy (abb0t):

Haha! I will work out the problem on paper tomorrow morning just to double-check and make sure that I didn't OK something that isn't right. But from what I see, it looks right :P Best of Luck with ODE's. Cheers!

OpenStudy (anonymous):

even i am getting the same answer............

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