## Calculus AB and Calculus BC

## CHAPTER 9 Differential Equations

### D. SOLVING FIRST-ORDER DIFFERENTIAL EQUATIONS ANALYTICALLY

In the preceding sections we solved differential equations graphically, using slope fields, and numerically, using Euler’s method. Both methods yield approximations. In this section we review how to solve some differential equations *exactly.*

**Separating Variables**

A first-order d.e. in *x* and *y* is *separable* if it can be written so that all the terms involving *y* are on one side and all the terms involving *x* are on the other.

A differential equation has variables separable if it is of the form

The general solution is

**EXAMPLE 8**

Solve the d.e. given the initial condition *y*(0) = 2.

**SOLUTION:** We rewrite the equation as *y* *dy* = *−x* *dx*. We then integrate, getting

Since *y*(0) = 2, we get 4 + 0 = *C*; the particular solution is therefore *x*^{2} + *y*^{2} = 4. (We need to specify above that *y* > 0. Why?)

**EXAMPLE 9**

If and *t* = 0 when *s* = 1, find *s* when *t* = 9.

**SOLUTION:** We separate variables:

then integration yields

Using *s* = 1 and *t* = 0, we get so *C* = + 2. Then

When *t* = 9, we find that *s*^{1/2} = 9 + 1, so *s* = 100.

**EXAMPLE 10**

If (ln *y*) and *y* = *e* when *x* = 1, find the value of *y* greater than 1 that corresponds to *x* = *e*^{4}.

**SOLUTION:** Separating, we get We integrate:

Using *y* = *e* when *x* = 1 yields so

When *x* = *e*^{4}, we have thus ln^{2} *y* = 9 and ln *y* = 3 (where we chose ln *y* > 0 because *y* > 1), so *y* = *e*^{3}.

**EXAMPLE 11**

Find the general solution of the differential equation

**SOLUTION:** We rewrite

Taking antiderivatives yields *e ^{u}* =

*e*+

^{v}*C*, or

*u*= ln(

*e*+

^{v}*c*).