Download 3.1 Integrating Both Sides

3.1
Integrating Both Sides
Implicit differentiation is a technique from calculus for taking the derivative of
expressions involving an unknown function y ( x ) . The idea is to use the normal
algorithm for differentiation, and to simply write y 0 whenever we would usually need
to take the derivative of y.
For example, to take the derivative of something like (sin x ) 3 , we would normally
use the power rule and the chain rule:
g
d f
(sin x ) 3 3 (sin x ) 2 cos x.
dx
Here the cos x comes from the chain rule,
since cos x is the derivative of sin x.
We use exactly the same procedure to take the derivative of something like y 3 , except
that we write y 0 whenever we need the derivative of y:
d f 3g
y 3y 2 y 0 .
dx
Here the 3y 2 comes from the product rule, and we multiply by y 0 because of the chain
rule.
EXAMPLE 1
Take the derivative of both sides of the following equation:
sin ( y ) + x 2 y x 3 .
SOLUTION
We can use the chain rule to take the derivative of sin ( y ) :
g
d f
sin ( y ) cos ( y ) y 0 .
dx
The derivative of x 2 y requires the product rule:
d f 2 g
x y 2x y + x 2 y 0 .
dx
Thus the derivative is
y 0 cos ( y ) + 2x y + x 2 y 0 3x 2 .
In general, the product rule states that
d f g
uv u 0 v + uv 0 .
dx
In this case, u x 2 and v y, so u 0 2x
and v 0 y 0 .
We can use implicit differentiation to find a differential equation that has a given
general solution. For example, suppose we want to find a differential equation whose
general solution is
y Cx 2 .
We start by solving for the constant C:
yx −2 C.
If we now take the derivative of both sides, the C disappears:
y 0 x −2 − 2yx −3 0.
This is a differential equation whose general solution is y Cx 2 . We can simplify the
equation a bit by multiplying through by x 3 :
x y 0 − 2y 0.
Since C is a constant, its derivative is zero.
INTEGRATING BOTH SIDES
2
EXAMPLE 2
Find a differential equation whose general solution is y ln ( x + C ) .
SOLUTION
To solve this equation for C, we first take the exponential of both sides:
e y x + C.
Thus
e y − x C.
Taking the derivative of both sides gives
e y y 0 − 1 0.
This is the differential equation we wanted. We can simplify it by solving for y 0 :
y 0 e −y .
Integrating Differential Equations
Now that we know how take derivatives implicitly, we can reverse the process to solve
differential equations. For example, given a differential equation like
2y y 0 4x 3 + 2,
Technically, there might be a constant on
both sides of the antiderivative, i.e.
we can integrate both sides to get an equation involving x and y:
y 2 x 4 + 2x + C.
y 2 + C1 x 4 + 2x + C2
However, we can combine the two
constants by subtracting C1 from both
sides of this equation.
We can now solve for y to get the general solution to the given differential equation:
p
y ± x 4 + 2x + C.
EXAMPLE 3
Find the general solution to the equation y 2 y 0 e 3x .
SOLUTION
We can integrate both sides to get
1
1 3
y e 3x + C.
3
3
Solving for y gives
y p3
e 3x + 3C.
Since C is an arbitrary constant, 3C is actually itself an arbitrary constant. If we let A 3C, we
can rewrite our general solution as
y where A is an arbitrary constant.
p3
e 3x + A,
INTEGRATING BOTH SIDES
3
EXAMPLE 4
Solve the following initial value problem:
e 4y y 0 x,
SOLUTION
y (0) 1.
In general, the antiderivative of f ( y ) y 0 is
Integrating both sides yields
Z
1 4y
1
e x 2 + C.
4
2
Though we could solve for y at this point, it is easier to start by substituting in the initial
condition:
1 4
e 0 + C.
4
1
Then C e 4 , so the solution becomes
4
f ( y ) dy.
In this case, the antiderivative of e 4y y 0 is
Z
e 4y dy 1 4y
e + C.
4
1
1
1 4y
e x2 + e 4 .
4
2
4
Solving for y gives
y 1 2
ln 2x + e 4 .
4
Sometimes it is possible to recognize one side of an equation as the result of the
product rule.
EXAMPLE 5
Find the general solution to the differential equation x 3 y 0 + 3xx 2 y 1.
The left side of this equation is the result of a product rule, with the two factors
being x 3 and y. Integrating both sides gives
SOLUTION
x 3 y x + C.
Solving for y yields the general solution:
y x −2 + Cx −3 .
A Closer Look Implicit Solutions
Sometimes after you integrate both sides of a differential equation it is not possible to
algebraically solve for y. For example, consider the equation
3y 2 + e y y 0 x 2 .
Integrating both sides gives
1 3
x + C.
3
Unfortunately, there is no way to solve this equation algebraically for y. This implicit solution
still describes the solutions of the equation, in the sense that the corresponding curves in
the x y-plane are graphs of the solution functions (see Figure 1), but there is no way to write
explicit algebraic formulas for these solutions.
y3 + e y a Figure 1: Three curves of the form
y3 + e y 1 3
x + C.
3
INTEGRATING BOTH SIDES
4
A Closer Look Integrating Second-Order Equations
The method of integrating both sides can also be used on second-order equations, although
the antiderivatives are often much more difficult. For example, consider the equation
y y 00 + ( y 0 ) 2 6x.
Though it may not be obvious, the left side is a result of the product rule: it is the derivative of
the product y y 0 . Integrating both sides gives
y y 0 3x 2 + C.
We have now reduced to a first-order equation, and we can integrate both sides again to obtain
the general solution. Specifically, we get
1 2
y x 3 + Cx + D,
2
where D is an arbitrary constant. Multiplying by two and taking the square root gives
p
y ± 2x 3 + Ax + B,
where A 2C and B 2D are arbitrary constants.
EXERCISES
Take the derivative of both sides of the given equation with respect to x.
1–2
1. x y + ln y sin (2x )
2. e 2y − x y 2 1
Find a differential equation whose general solution is the given formula.
3–6
√
√
3. y C x
4. y ± x + C
5. y tan ( x + C )
6. y e 2x + Ce −x
Find the general solution to the given differential equation by integrating both
7–12
sides.
7. y y 0 e x
√
9. y 0 y cos x
11. e 3x y 0 + 3e 3x y e 5x
13–14
8. y 0 e 2y x 3
10. x 4 y 0 + 4x 3 y x 4
12. y 0 tan x + y sec2 x cos x
Solve the given initial value problem.
13. y 0 cos y e −x , y (0) 0
14. −y −2 y 0 2x, y (0) 3