2.4 Linear Equations
(f g)0 = f g 0 + f 0 g
Product rule:
x = x(t), (tx)0 = tx0 + x. On the other hand tx0 + x = (tx)0
A few more examples:
Solve the equation: tx0 + 2x = 4
Example:
Solution:
t2 x0 + 2tx = (t2 x)0 , et x0 + et x = (et x)0 , sin tx0 + cos tx = (sin tx)0
t2 x0 + 2tx = 4t,
(t2 x)0 = 4t,
R
(t2 x)0 dt =
R
4t dt,
t2 x = 2t2 + C
x(t) = 2 + Ct−2
First Order Linear Equation:
b(t)x0 + c(t)x + g(t) = 0 is a general form of a first order linear differential equation.
It is linear with respect to both x0 and x.
[Quasilinear equation: b(t, x) x0 + c(t, x) = 0. It is linear with respect to x0 .]
b(t) x0 = −c(t) x − g(t)
After division by b(t) we convert it to x0 = a(t) x + f (t)
which is an inhomogeneous equation due to the presence of f (t).
Homogeneous equation:
The equation
x0 = a(t) x
is homogeneous. A homogeneous first order linear equation is a separable equation.
Solution:
R dx R
dx
= a(t)x ⇒
= a(t)dt
dt
x
R
R
ln |x| = a(t)dt + c ⇒ x(t) = Ae a(t) dt (after exponentiating)
Example:
R dx R
= t dt,
x
x0 = tx
ln |x| =
t2
+ c,
2
2 /2
x(t) = Aet
1
Example:
tx0 = x
R dx R dt
=
,
x
t
Example:
x(t) = At
tx3 x0 = 1
The equation is nonlinear but separable
R
x3 dx =
R dt
,
t
x4
= ln |t| + c1 ,
4
1
x = ±(4 ln |t| + c) 4
Alternative solution:
dt
= x3 t
dx
We change the equation and make it linear: tx3 dx = dt,
The last equation linear with respect to the dependent variable t as a function of the independent
variable x.
Inhomogeneous equation:
Now let us consider an inhomogeneous first order linear equation
x0 = a(t)x + f (t)
(1)
It can be solved by two different methods.
Method 1. Integrating Factor
To solve the equation we apply the following steps:
1. Rewrite the equation in the form x0 − a(t) x = f (t).
2. Multiply it by the integrating factor
u(t) = e−
R
a(t)dt
which is solution to the equation u0 = −a(t) u to get
u(x0 − ax) = ux0 − aux = ux0 + u0 x = (ux)0 = uf .
3. Integrate (ux)0 = uf to obtain ux =
R
u(t)f (t)dt + c.
4. Solve the last equation for x:
1
x(t) =
u(t)
Z
u(t)f (t) dt +
2
c
u(t)
Method 2. Variation of Parameters
1. Find a particular solution xh (t) to the corresponding homogeneous equation x0 = ax,
xh (t) = e
R
a(t)dt
.
f
.
xh
Indeed, after substitution x = v xh into (1) we get v 0 xh + vx0h = avxh + f
2. Substitute x = v xh into (1) to find v(t) or remember that v 0 =
Recalling that x0 = ax we obtain v 0 xh = f and finally v 0 =
f
.
xh
Integrate the last equation and get v(t).
3. Then the general solution is x(t) = v(t) xh (t)
Example:
x0 + x = t
a = −1
The integrating factor is u(t) = e
R
1dt
= et
After multiplying both sides by u(t) = et we get et x0 + et x = et · t or (et x)0 = tet .
Then et x =
R
tet dt = tet − et + c (integration by parts)
x(t) = (tet − et + c) e−t ,
Check: x0 = 1 − ce−t ,
x(t) = t − 1 + ce−t is the solution.
x0 + x = 1 − ce−t + t − 1 + ce−t = t. Correct.
Method 2 (variation of parameters):
The corresponding homogeneous equation is x0 + x = 0.
Its solution is xh = cs−t that we write in the form xh = v(t)e−t .
Substitution into inhomogeneous equation gives (ve−t )0 + ve−t = t,
v 0 e−t − ve−t + ve−t = t,
v 0 e−t = t,
v 0 = tet ,
Therefore, x(t) = v(t)e−t = t − 1 + ce−t .
3
v(t) = tet − et + c
tx0 = 2x + t3
Example:
Linear equation is x0 =
2
x + t2 , (t 6= 0).
t
Method 1
u = e−
R
2
dt
t
2
= e−2 ln(t) = e− ln t = t−2
t−2 x0 = 2t−3 x + 1,
t−2 x0 − 2t−3 x = 1,
(t−2 x)0 = 1,
t−2 x = t + c,
x(t) = t3 + ct2
Case t = 0. If t = 0 then x(0) = 0 and the equation holds, hence x(t) defined on (−∞, ∞).
Method 2
The corresponding homogeneous equation of the linear equation is x0 =
R dx R dt
= 2 ,
x
t
ln |x| = ln t2 + c,
x0 = v 0 · t2 + 2vt =
Example:
2
· vt2 + t2 ,
t
2
x = eln t ec = At2 ,
v 0 t2 = t2 ,
v 0 = 1,
2
x
t
x = v(t) t2
v = t + c,
x(t) = t3 + ct2
x = (2t + x3 )x0
The equation is linear with respect to t, not x
x = (2t + x3 )
dx
,
dt
dt · x = (2t + x3 )dx,
dt
· x = 2t + x3
dx
So we need to find a solution t(x) to the equation xt0 = 2t + x3 .
It is exactly the previous example with t and x switched.
Therefore, the solution is t = x3 + cx2 . It gives us an implicit solution in terms of x(t).
4