5 Steps to a 5 AP Calculus BC 2019

More Applications of Definite Integrals 323

Step 6. Separate variables:dy=(x^2 +x−1)dx.

Step 7. Integrate both sides:

∫

dy=

∫

(x^2 +x−1)dx

y=

x^3

3

+

x^2

2

−x+C 2.

Step 8. Substitute given condition: Atx=0,y=3; 3= 0 + 0 − 0 +C 2 ⇒C 2 =3.

Therefore,y=

x^3

3

+

x^2

2

−x+ 3.

Step 9. Verify the result by differentiating:

y=

x^3

3

+

x^2

2

−x+ 3

dy

dx

=x^2 +x−1;

d^2 y

dx^2

= 2 x+ 1.

Example 4

Find the general solution of the differential equation
dy
dx

=

2 xy

x^2 + 1

.

Step 1. Separate variables:

dy

y

=

2 x

x^2 + 1

dx.

Step 2. Integrate both sides:

∫

dy

y

=

∫

2 x

x^2 + 1

dx(letu=x^2 +1;du= 2 xdx)

ln|y|=ln(x^2 +1)+C 1.

Step 3. General Solution: solve fory.

eln|y|=eln(x^2 +1)+C^1

|y|=eln(x^2 +1)·eC^1 ;|y|=eC^1 (x^2 +1)

y=±eC^1 (x^2 +1)

The general solution isy=C(x^2 +1).

Step 4. Verify the result by differentiating:

y=C(x^2 +1)

dy

dx

= 2 Cx= 2 x

C(x^2 +1)

x^2 + 1

=

2 xy

x^2 + 1

.