Exact differential equation

An exact differential equation is a differential equation that can be solved in the following manner.

Suppose you are given an equation of the form:

$M(t,y)+N(t,y)y'=0\,$

or

$M(t,y)dt+N(t,y)dy=0\,$

To find the solution of this equation, we assume that the solution is φ = constant. This means that: ${\frac {\partial \phi }{\partial t}}=M$ and ${\frac {\partial \phi }{\partial y}}=N$

φ is found by integrating M and N:

$\phi (t,y)=\int _{0}^{t}M(s,0)ds+\int _{0}^{y}N(t,s)ds$

Go through the example to find φ by integrating, then check that

${\frac {\partial \phi }{\partial t}}=M$

and

${\frac {\partial \phi }{\partial y}}=N$

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

${\frac {\partial M}{\partial y}}={\frac {\partial N}{\partial t}}$

(That's the condition for "exactness" of the differential form M dt + N dy.)

where ${\frac {\partial M}{\partial y}}={\frac {\partial N}{\partial t}}$.

To find $y$, first set $M={\frac {\partial \phi }{\partial t}}$ and $N={\frac {\partial \phi }{\partial y}}$. Then manipulate to get $M\partial t=\partial \phi$ and $N\partial y=\partial \phi$. Integrate both sides, compare the results for $\phi$, 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 $y$, plug into the quadratic formula.