So far, the only general method we have for solving differential equations involves equations of the form
y0 = f(x),
where f(x) is any function of x. The solution to such an equation is obtained by integrating:
Of course, we also know how solve equations of the formy00=f(x) by
integrating twice. y =
Z
f(x)dx
In this section, we will expand upon this method by learning how tointegrate both sides of an equation. This will allow us to solve many different types of differential equations.
Implicit Differentiation
The method of integrating both sides is related to a technique from calculus known as implicit differentiation. This is the procedure we use to take the derivative (with respect to x) of an expression that involves both xandy.
The idea of implicit differentiation is to imagine thaty represents some unknown formula, and to simply writey0 whenever we would usually write the derivative of that formula.
EXAMPLE 1 Compute the derivative with respect to xof each of the following ex-pressions.
(a) y3 (b) x2y (c) x3e2y
SOLUTION
(a) Imagine that y represents some unknown formula, such as sinx. In this case,
y3would be (sinx)3, and we could take its derivative using the chain rule:
d dx
(sinx)3 = 3(sinx)2cosx.
In the same way, we can use the chain rule to take the derivative ofy3:
d dx
y3 = 3y2y0
wherey0 represents the derivative ofy, whatever that may be.
(b) Again, to take the derivative of something likex2sinx, we would normally use the product rule:
d dx
x2sinx
= x2cosx+ 2xsinx.
Therefore, we must also use the product rule to take the derivative ofx2y:
d dx
x2y
= x2y0+ 2xy
(c) Here we must use a combination of the product rule and the chain rule:
d dx
When we are given an equation involvingxandy, we can use implicit differentiation to take the derivative of both sides.
EXAMPLE 2 Take the derivative of both sides of each equation with respect tox. (a) y2+x4 = e3x (b) x3y = 4 sinx.
SOLUTION The solutions are
2yy0+ 4x3= 3e3x and x3y0+ 3x2y= 4 cosx
Integrating Both Sides
The idea of integrating both sides is to reverse this process. Given a differential equation like
3y2y0 = 4x3+ 2,
we take the antiderivativeof both sides to get an equation involvingxandy: Technically, there might be a constant
on both sides of the antiderivative, i.e. y3+C
1=x4+ 2x+C2 However, we can combine the two constants by subtractingC1 from both sides of this equation.
y3 = x4+ 2x+C.
We can now solve fory to get the general solution to the given differential equation:
y = p3
x4+ 2x+C.
EXAMPLE 3 Find the general solution to each of the following differential equations. (a) y2y0 =e2x (b) x3y0+ 3x2y= 1 (c) 2yy0lnx+y
2
x = 0 SOLUTION
(a) We can take the antiderivative of both sides to get 1
3y
3= 1
2e
2x+C.
Solving fory yieldsy= 3
r
3 2e
2x+ 3C, or equivalently y= 3
r
3 2e
2x+C . Here we have replaced 3CbyC, since
three times an arbitrary constant is just an arbitrary constant. Note that value ofCchangedwhen we did this.
(b) The left side of this equation looks like a result of the product rule. Therefore, the antiderivative is
x3y=x+C.
Solving fory gives y=x−2+Cx−3 .
(c) Again, the left side is a result of the product rule, with the two factors being y2
and lnx. Therefore, the antiderivative is
y2lnx=C.
Solving fory givesy=±
r
C
lnx, or equivalently y= C
√
lnx .
Here we pulled theCoutside of the square root and combined it with the±. Again, this changes the value
ofC, but the result is equivalent. Sometimes it necessary to manipulate an equation algebraically before it can be integrated.
Implicit Solutions
Sometimes after you integrate both sides of a differential equation it is not possible to algebraically solve fory. For example, consider the equation
3y2y0+eyy0 = x2.
Integrating both sides gives
y3+ey = 1 3x
3
+C.
Unfortunately, there is no way to solve this equation algebraically for C. This implicit solutionstill describesy as a function ofx, in the sense that the curves in thexy-plane defined by this equation are graphs of the solutions. However, there is no way to write a formula for these solutions algebraically.
EXAMPLE 4 Find the general solution to each of the following differential equations. (a) y0= cosx
4y3 (b) x
3y0=x5−3x2y
SOLUTION
(a) The right side of this equation cannot be integrated, since it has y but not y0. However, we can make this equation integrable if we multiply through by 4y3:
4y3y0 = cosx.
Integrating both sides gives
y4= sinx+C Note that you need to include a±
when you take the fourth root of both sides of an equation, just like for the square root. In general, the±is necessary for thenth root whenevern is even.
so the general solution is y=±√4
sinx+C .
(b) This time, neither side of the the equation can be integrated. However, we can make this equation integrable if we move the 3x2y to the left side:
x3y0+ 3x2y=x5.
The left side is now a result of the product rule. Integrating both sides gives
x3y= 1 6x
6+C
so the general solution is y= 1 6x
3+Cx−3 .
Separation of Variables
For many equations, the best way to make the equation integrable is toseparate the variables so that all the x’s lie on one side of the equation, and all of the y’s lie on the other side. For example, consider the equation
−y0 = 3x2ey.
To make this equation integrable, the best strategy is to divide through by ey,
sepa-rating the two variables:
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
yy00+ (y0)2 = 6x.
Though it may not be obvious, the left side is a result of the product rule: it is the derivative of the productyy0. Integrating both sides gives
yy0= 3x2+C1.
We have now reduced to a first-order equation, and integrating both sides again yields 1
2y
2
=x3+C1x+C2
Therefore, the general solution isy=±p2x3+C
1x+C2.
Integrating both sides yields
e−y = x2+C
and solving for ygivesy=−ln x2+C.
In general, if you can put an equation into the form
f(y)y0 = g(x), A differential equation that can be put
into this form is calledseparable.
then you can integrate both sides by computing a pair of integrals.
SEPARATION OF VARIABLES
Given any equation of the form
f(y)y0 = g(x),
we can integrate both sides to get
F(y) = G(x) +C
whereF(y) =
Z
f(y)dy, and G(x) =
Z
g(x)dx.
EXAMPLE 5 Find the general solution to each of the following differential equations. (a) y0ex=xy2 (b) xy0= cos2y
SOLUTION
(a) To separate the variables, we must divide through by both exandy2. This gives
y−2y0 = xe−x
We can now integrate both sides, using integration by parts for
Z
xe−xdx. The result is
−y−1 = −(x+ 1)e−x+C. As we have seen before, this general
solution misses one possibility: y= 0 is a solution, but it does not correspond to any value ofC. (Intuitively,y= 0 is the solution corresponding toC=∞.)
Solving fory givesy= 1
(x+ 1)e−x−C or equivalently y=
1
Separation of Variables using Differentials
There is a nice way of viewing separation of variables using differentials, which looks a little bit different from our method. Given an equation like
exy0 =xy3,
we can write it as
exdy dx = xy
3
,
and then “separate variables” as follows:
y−3dy = xe−xdx. Actually, the differentialsdxanddy
representinfinitesimallysmall changes inxandy. Infinitesimals are not often used in mathematics, since they are not part of the real number system, but reasoning using infinitesimals is common in the sciences.
Here, thedxanddyby themselves are “differentials”, which represent small changes in the values ofxandy. Then
Z
y−3dy = Z
xe−xdx,
and evaluating these two integrals gives the solution.
(b) To separate variables, we must divide through by bothxand cos2y. This gives
sec2y
y0 = 1
x Recall that
Z
sec2y dy = tany+C
and Z
1
xdx = ln|x|+C. The absolute value in the integral of 1/xis important: it makes it possible forxto be negative.
We can now integrate both sides, which yields tany = ln|x|+C,
so the general solution is y= tan−1 ln|x|+C.
EXAMPLE 6 Find the general solution to the equation
y0 = ycosx.
SOLUTION We can divide through byy to separate variables:
y0
y = cosx.
Integrating both sides yields
ln|y| = sinx+C.
To solve fory, we must take the exponential of both sides. This gives
|y| = esinx+C,
which is the same as
y = ±esinx+C.
There is a complicated way to simplify this formula. First, we split up the exponential: Pay careful attention to the trick used
here.
y = ±eCesinx.
Now, since C is an arbitrary constant, eC may be any positive constant. Then ±eC
may be positive or negative, so it is essentially an arbitrary constant. Therefore, we can write the general solution as
The trick used at the end of this example is fairly common. Given a general solution of the form
|y| = ef(x)+C
we can use this trick to rewrite the solution as
y = Cef(x).
Integrating Factors
Consider the following differential equation:
xy0+ 3y=x4.
This equation cannot be integrated directly, and it is not obvious how to perform algebra to make it integrable. In particular, it certainly won’t work to separate the variables.
There is actually a very clever trick we can use to make this equation integrable. We need tomultiply through by x2:
x3y0+ 3x2y=x6.
As you can see, the left side is now a result of the product rule. Integrating both sides gives
x3y = 1 7x
7+C
so y=1 7x
4+Cx−3.
The factor ofx2that we multiplied by is called aintegrating factor—a factor that
you multiply to make an equation integrable. There is a well-known class of differential equations for which integrating factors can be quite helpful.
FIRST-ORDER LINEAR EQUATIONS
A differential equation of the form A differential equation of the form
f(x)y00 +g(x)y0 +h(x)y = k(x). is called asecond-order linear equation. We will be discussing such equations later in the course.
f(x)y0 +g(x)y = h(x)
is called afirst-order linear equation. Such an equation is integrable if
f0(x) = g(x).
It turns out thatany first-order linear equation can be made integrable if you multiply by the appropriate integrating factor. When searching for this factor, keep in mind that the goal is to make the derivative off(x) equal tog(x).
EXAMPLE 7 Find the general solution to each of the following differential equations. (a) x2y0+ 6xy=x+ 1 (b) y0+ 2y=x (c) y0−ytanx= 1
SOLUTION All of these equations are linear, so we ought to be able to find integrating factors.
(a) This equation isn’t integrable, since 6xisn’t the derivative ofx2. However, we can make it integrable if we multiply through byx4:
The Integrating Factor Formula
There is actually a formula for the integrating factor required to make a first-order linear equation integrable. Given the equation
y0+g(x)y=h(x)
the appropriate integrating factor is eG(x), whereGis any antiderivative of g. Note that any first-order linear equation can be brought into this form by dividing though by the coefficient ofy0.
Although this formula works for any first-order linear equation, it can be hard to remember, and the algebra is often unnecessarily complicated. For simple equations, it is usually much easier to guess the integrating factor and then work from there.
The left side is now the derivative ofx6y. Integrating both sides gives
x6y= 1 6x
6+1
5x
5+C
Solving fory gives y= 1 6+
1 5x
−1+Cx−6.
(b) Multiplying by a power ofxwon’t work here, since 2xa is never the derivative of xa. Instead, we need to multiply through bye2x:
e2xy0 + 2e2xy = xe2x.
The left side is now the derivative ofe2xy. We can now integrate both sides, using
integration by parts on the right side.
e2xy = 1 2xe
2x−1
4e
2x+C.
Solving fory gives y= 1 2x−
1 4 +Ce
−2x .
(c) A little bit of guessing and checking is required to find an integrating factor for this equation. The right strategy is to multiply through by cosx:
y0cosx−ysinx= cosx.
The left side is the derivative ofycosx. Integrating both sides gives
ycosx= sinx+C
EXERCISES
1–6 Find the general solution to the given differential equation by integrating both sides. You may need to use some algebra to put the equation into integrable form.
1. 2yy0=ex 2. y0cosy=x3
3. x4y0+ 4x3y=x4 4. e3xy0+ 3e3xy=e5x 5. x2y0 = x−2xy 6. (y+y0)ex= sec2x 7–14 Use separation of variables to find the general solution to the given differential equation.
7. y0=xsecy 8. e−xy0=y2
9. ex+yy0= 1 10. (x2+ 1)y0=e−y 11. x2y0= 2xy2+y0 12. y0−y2= 1−x2y0
13. y0cos2x=y 14. xy0= 2y
15–22 Use an integrating factor to find the general solution to the given differential equation.
15. x2y0+ 3xy= 1 16. xy0−2y=x5
17. y0+ 5y=ex 18. y0−y=x
19. y0+ycotx= cosx 20. y0sinx+ysecx= cosx 21. xy0lnx+y=x2 22. y0+ 2xy=x
23–24 Solve the given initial value problem.
23. y0=−3x2y2,y(0) = 3 24. y0=x/y,y(3) =−5
1.y=±√ex+C 2.y=sin−1
x4
4 +C
3.y=x
5+
C
x4 4.y=
1 5e
2x+Ce−3x 5.y=1 2+Cx
−2
6.y=e−x(tanx+C) 7.y=sin−1
1
2x
2+C
8.y= 1
C−ex 9.y=ln c−e −x
10.y=ln arctanx+C
11.y= 1
C−ln(x2−1) 12.y=tan arctanx+C
13.y=Cetanx 14.y=Cx2 15.y= 1
2x+ C x3
16.y=x
5
3 +Cx
2 17.y=ex
6 +Ce
−5x 18.y=Cex−x−1.
19.y=1
2sinx+Ccscx or y=− 1
2cosxcotx+Ccscx. 20.y= (x+C)cotx. 21.y=
x2+C
2 lnx
22.y=1
2+Ce
−x2 23.y= 3
3x3+1 24.y=−
p
x2+16 25.y=x 4+3
4x2 26.y=
9e3x−ex 2
