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{\Large\bf Separable First-Order Differential Equations}

\vspace{.2in}

A separable integrable first-order ode has the form:

$$
y'(t)=g(t)h(y) \hspace{.2in} y(t_0)=y_0
$$

where g(t) and h(y) are given functions. Note that
y'(t) is the product of functions of the independent variable
and dependent variable.

\vspace{.3in}

{\bf Solution Procedure}

\vspace{.2in}

To solve this problem, divide by h(y):

$$
y'(t)/h(y(t))=H(y(t))y'(t)=g(t)
$$

Here we have relabeled 1/h(y(t)) by H(y(t)). Now integrate
both sides with respect to t 

$$
\int H(y(t))y'(t)dt=\int g(t)dt
$$

The left-hand side can be simplified using the method of substition
for integrals. Let u=y(t). Then du=y'(t)dt. We then have

$$
\int H(y(t))y'(t)dt= \int H(u)du 
$$

The final answer is (we change from the dummy variable u to
the dummy variable y)

$$
\int H(y)dy = \int g(t)dt +C
$$

We can also write the answer in terms of definite integrals
if an initial condition is given. We have

$$
\int_{t_0}^{t}H(y(s))y'(s)ds=\int_{t_0}^{t}g(s)ds
$$

Make the subsitution in the first integral u=y(s), du=y'(s)ds.
The lower limit of the new integral is 

$$
y(t_0)=y_0.
$$

The upper limit of the new integral is y(t). We then have

$$
\int_{y_0}^{y(t)}H(u)du=\int_{t_0}^{t}g(s)ds
$$

\vspace{.3in}

{\bf Example}

\vspace{.2in}

ODE:

$$
y'=(exp(t)+1)/y \hspace{.2in} y(1)=3
$$

Separating we have:

$$
yy'=exp(t)+1
$$

In this case H(y)=y and g(t)=exp(t)+1. Computing the integrals
we have

$$
\int H(y) dy=\int ydy= y^2/2
$$

$$
\int g(t)dt= \int (exp(t)+1)dt=-exp(t)+t
$$

The general solution of the ode is

$$
y^2/2=exp(t)+t+C
$$

Applying the initial condition, we find that 

$$
3^2/2=exp(1)+1+C
$$

C=9/2-exp(1)-1=0.782. The solution of the ode is

$$
y^2/2=exp(t)+t+0.782
$$

We can solve for y. We obtain

$$
y(t)=\sqrt{2exp(t)+2t+1.564}
$$

Notice that we take the positive square root. Why? We know
that y(1)=3, a positive number. 

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