Separable First-Order ODE

A separable first-order ode has the form:

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.

Solution Procedure

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

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

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

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

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

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

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

Example

ODE:

Separating we have:

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

The general solution of the ode is

Applying the initial condition, we find that

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

We can solve for y. We obtain

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

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