Question

First order exact ode

Answer 1

\[M+Ny'=0\]\[M\text dx+N\text dy=0\]\[MR(y)\text dx+NR(y)\text dy=0\]\[\frac{\partial MR(y)}{\partial y}=\frac{\partial NR(y)}{\partial x}\]\[\frac{\partial MR(y)}{\partial y}=R(y)\frac{\partial N}{\partial x}\]\[M\frac{\partial R(y)}{\partial y}+R(y)M_y=R(y)N_x\]\[M\frac{\partial R(y)}{\partial y}+R(y)\left(M_y-N_x\right)=0\]\[R(y)'+R(y)\frac{\left(M_y-N_x\right)}{M}=0\]\[R(y)=e^{\int\limits^y\frac{\left(M_y-N_x\right)}{M}\text dy}=e^{\int\limits^y Q(t)\text dt}\]

Answer 2

either the question or myself seams to have something backwards

Answer 3

or is \[Q(t)=\frac{M_y-N_x}{M}=0\]?

Answer 4

/????

Answer 5

i dont know if i have solved the question or not

Answer 6

You are making mistake
the solution to the equation
\[ y' + k y = 0 \text { is } y=c_1 e^{-\int k dx }\]
assuming c1 to be constant, you have your IF

Answer 7

*assuming c1 = 1, you have your IF

Answer 8

my text dosent have a minus in the integrating factor , can you direct me to a text that does?

Answer 9

\[ R(y)'+R(y)\frac{\left(M_y-N_x\right)}{M}=0 \]
\[ R(y) = C e^{-\int \frac{\left(M_y-N_x\right)}{M} dx } = C e^{\int \frac{-\left(M_y+N_x\right)}{M} dx } \]

Answer 10

\[ R(y) = C e^{-\int \frac{\left(M_y-N_x\right)}{M} dx } = C e^{\int \frac{N_x - M_y}{M} dx } \]

Answer 11

*error in previous post

Answer 12

i do not understand

Answer 13

where does the negative sign come from?

Answer 14

Linear ODE of form \( y' + ky = 0 \) has solution of the form \( y (IF) = constant \)
or \( y = c(IF)^{-1}\) ... and put c=1

Answer 15

oh, ok.