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{\Large\bf Exact Equations}

\vspace{.2in}

An exact equation has the form:

$$
f(t,y)dt+g(t,y)dy=0 \hspace{.2in} (*)
$$

where the partial derivatives of f and g satisfy

$$
f_y=g_t \hspace{.2in} (*)
$$

\vspace{.3in}

{\bf Solution Procedure}

The first step for solving this problem is to relate the
ode to a differential in two dimensions.  For a function of
two variables, Z(t,y), the differential is

$$
dZ=Z_tdt+Z_ydy
$$

If we set the differential equal to 0, we have dZ=0. If the
differential is 0, then the function is a constant. The
solution to dZ=0 is Z(t,y)=constant. This defines a family of
curves in the t-y plane. 

Let's go back to the differential equation (*). If we
can make the connection that f(t,y) is the t derivative
of a function Z(t,y) and g(t,y) is the y derivative of Z(t,y)
then the equation says dZ=0, implying Z(t,y)=constant is
the solution. We need only find Z(t,y) and we are done.

Equation (**) is a necessary condition for ode (*) to be exact.
Why is (**) necessary? Recall that if 

$$
Z_t=f \hspace{1in} Z_y=g,
$$

then the mixed partial second derivatives satisfy 

$$
Z_{ty}=Z_{yt},
$$

if the derivatives are continuous.

Suppose now that we know that

$$
Z_t=f \hspace{1in} Z_y=g
$$

where f and g are given and that the necessary condition (**)
is satisfied. How do we find Z(t,y)? By integration! By the
Fundamental Theorem of Calculus, we have

$$
Z(t,y)=\int f(t,y)dt+C(y)
$$

Here we have added a constant which is a function of y, since
we are integrating with respect to t. We also have

$$
Z(t,y)=\int g(t,y)dy+D(t)
$$

Now the constant is a function of t. We find C(y) and D(t)
by requiring 

$$
\int f(t,y)dy+C(y)=\int g(t,y)dy+D(t)
$$

The final solution is

$$
Z(t,y)= \int f(t,y)dy+C(y)=\int g(t,y)dy+D(t)=E,
$$

where E is a constant.

\vspace{.3in}

{\bf Example}

\vspace{.2in}

ODE:

$$
(y\cos(t)+2texp(y))dt+(\sin t + t^2exp(y)+2)dy=0
$$

First, we check if the formula is exact. We have

$$
f(t,y)= y\cos(t)+2texp(y) \hspace{.2in}\Longrightarrow
\hspace{.2in} f_y=\cos(t)+2texp(y)
$$

$$
g(t,y)= \sin t + t^2exp(y)+2
\hspace{.2in}\Longrightarrow\hspace{.2in}
g_t=\cos t +2texp(y)
$$

Since the two partial derivatives are equal (and continuous)
the equation is exact.

Integrating with respect to t, we have 

$$
Z(t,y)=\int(y\cos(t)+2texp(y))dt+C(y)=y\sin(t)+t^2exp(y)+C(y)
$$

Integrating with respect to y, we have

$$
Z(t,y)=\int(\sin t+t^2exp(y)+2)dt+D(t)=y\sin(t)+t^2exp(y)+2y+D(t)
$$

Comparing these two expressions we conclude

$$
C(y)=2y+D(t)
$$

Choose C(y)=2y and D(t)=0. This isn't the only 
solution. We could choose C(y)=2y+10 and D(t)=10, for example.

The final solution to the ode is 

$$
Z(t,y)=y\sin(t)+t^2exp(y)+2y=constant
$$

\end{document}
