Changes
:<math>M(t,y) + N(t,y)y' = 0\,</math> or <math>M(t,y) dt + N(t,y) dy = 0\,</math>
<math>\frac{\partial M}{\partial y} = \frac{\partial N}{\partial t}</math> To find the solution of this equation, we assume that the solution is φ is found = constant. We can re-write a different form of this equation by integrating substituting <math>\frac{\partial \phi}{\partial t} = M </math> and N::<math>\frac{\partial \phi(t, }{\partial y) } = N</math>. This yields <math>(\int_0^frac{\partial \phi}{\partial t M(s, 0}) ds dt + (\frac{\partial \phi}{\int_0^partial y N(t, s}) dsdy = 0</math>, which makes sense. to find φ, we integrate M with respect to t and N with respect to y. This will give us two different equations. To find φ , we
Go through the example to find φ by integrating, then check that
and that any function φ = some constant, when turned into the corresponding dy/dt, satisfies the original equation. Be sure to emphasize that one must check first that
:<math>\frac{\partial M}{\partial y} = \frac{\partial N}{\partial t}</math>
To find <math>y</math>, first set <math>M = \frac{\partial \phi}{\partial t}</math> and <math>N = \frac{\partial \phi}{\partial y}</math>. Then manipulate to get <math>M \partial t = \partial \phi</math> and <math>N \partial y = \partial \phi</math>. Integrate both sides, compare the results for <math>\phi</math>, and combine the terms into one equation (for terms that show up in both expressions, only write once in the combined expression.) To solve the expression for <math>y</math>, plug into the quadratic formula.
