I like that u inject some humors even the problems get worst 😂..I learned a lot from u sir in unique way..hoping you'll be doing lessons on engineering mechanics.. God blessed sir
Its easy to prove that. A separable first order DE has the form M(x) dx + N(y) dy = 0. To test for exactness, get partial of M wrt y. You will get zero because M is purely a function of x. Solve for the partial of N wrt x, the answer is also zero because N is a function of y alone. Since the partial derivatives are equal (both zero) the equation is exact. :)
I like that u inject some humors even the problems get worst 😂..I learned a lot from u sir in unique way..hoping you'll be doing lessons on engineering mechanics.. God blessed sir
Thank you sir! 🤍
You're welcome! Keep on learning! 😊
It is true po ba na every separable first order in the form of dy/dx=g(x) h(y) is exact???? How can I prove though?
Its easy to prove that. A separable first order DE has the form M(x) dx + N(y) dy = 0. To test for exactness, get partial of M wrt y. You will get zero because M is purely a function of x. Solve for the partial of N wrt x, the answer is also zero because N is a function of y alone. Since the partial derivatives are equal (both zero) the equation is exact. :)
@@JefrilAmboy thank you po, big help 😊
na cot daw si x. XD
😄😄😄