Linear first-order differential equations

P dx =eln x = x (from the definition of logarithm).

(iv) Substituting into equation (3) gives:

yx = ∫

x(−1)dx

(v) Hence the general solution is:

yx = −x 2

2 +c

x −1 2

+c, from which, c= 3

2 Hence the particular solution is:

yx = −x 2

2 + 3

2 i.e. 2yx = 3− x2 and y = 3 − x

2x

Problem 3. Determine the particular solution of dy dx

− x + y=0, given that x =0 when y=2

Using the procedure of Section 79.2:

(i) Rearranging gives dy dx

+ y= x , which is of the

form dy dx

+ P,= Q, where P =1 and Q= x . (In this case P can be considered to be 1x0, i.e. a function of x)

(ii) ∫ P dx = ∫ 1dx= x (iii) Integrating factor e

(iv) Substituting in equation (3) gives:

yex = ∫

ex(x)dx (4)

(v) ∫ ex (x)dx is determined using integration by parts (see Chapter 68).∫

xex dx = xex − ex + c

Hence from equation (4): yex = xex −ex +c, which is the general solution. When x =0, y=2 thus 2e0=0−e0+c, from which, c=3 Hence the particular solution is:

yex = xex − ex + 3 or y = x− 1+ 3e−x

Now try the following Practice Exercise

Practice Exercise 301 Further problems on linear first-order differential equations (answers on page 1144)

Solve the following differential equations.