differential equation

456 DIFFERENTIAL EQUATIONS
equation. Given boundary conditions, the particular
solution may be determined.
(ii)
Q =−1. (Note that Q can be considered to be
−1x 0 , i.e. a function of x).
∫ ∫ 1
P dx = dx = ln x.
x
48.3 Worked problems on linear first
order differential equations
Problem 1. Solve 1 dy
+ 4y = 2 given the
x dx
boundary conditions x = 0 when y = 4.
Using the above procedure:
(i) Rearranging gives
dy + 4xy = 2x, which is
dx
of the form dy + Py = Q where P = 4x and
dx
Q = 2x.
(ii) ∫ Pdx = ∫ 4xdx = 2x 2 .
(iii) Integrating factor e ∫ P dx = e 2x2 .
(iv) Substituting into equation (3) gives:
∫
ye 2x2 = e 2x2 (2x)dx
(v) Hence the general solution is:
ye 2x2 = 1 2 e2x2 + c,
by using the substitution u = 2x 2 When x = 0,
y = 4, thus 4e 0 =
2 1 e0 + c, from which, c =
2 7 .
Hence the particular solution is
ye 2x2 = 1 2 e2x2 + 7 2
or y = 1 2 + 7 2 e−2x2 or y = 1 2
(1 + 7e −2x2)
Problem 2. Show that the solution of the equation
dy
dx + 1 =−y 3 − x2
is given by y = ,given
x 2x
x = 1 when y = 1.
Using the procedure of Section 48.2:
(i) Rearranging gives: dy
dx + ( 1
x
)
y =−1, which
is of the form dy
dx + Py = Q, where P = 1 x and
(iii) Integrating factor e ∫ P dx = e ln x = x (from the
definition of logarithm).
(iv) Substituting into equation (3) gives:
∫
yx = x(−1) dx
(v) Hence the general solution is:
yx = −x2
2 + c
When x = 1, y = 1, thus 1 = −1
2
which, c = 3 2
Hence the particular solution is:
i.e.
yx = −x2
2 + 3 2
2yx = 3 − x 2 and y = 3 − x2
2x
+ c, from
Problem 3. Determine the particular solution
of dy − x + y = 0, given that x = 0 when y = 2.
dx
Using the procedure of Section 48.2:
(i) Rearranging gives dy + y = x, which is of the
dx
form dy + P, = Q, where P = 1 and Q = x. (In
dx
this case P can be considered to be 1x 0 , i.e. a
function of x).
(ii) ∫ P dx = ∫ 1dx = x.
(iii) Integrating factor e ∫ P dx = e x .
(iv) Substituting in equation (3) gives:
∫
ye x = e x (x)dx (4)
(v) ∫ e x (x)dx is determined using integration by
parts (see Chapter 43).
∫
xe x dx = xe x − e x + c