Higher Engineering Mathematics

LINEAR FIRST ORDER DIFFERENTIAL EQUATIONS 457

I

Hence from equation (4):yex=xex−ex+c,

which is the general solution.

Whenx=0,y=2 thus 2e^0 = 0 −e^0 +c, from

which,c=3.

Hence the particular solution is:

yex=xex−ex+ 3 ory=x− 1 +3e−x

Now try the following exercise.

Exercise 183 Further problems on linear

first order differential equations

Solve the following differential equations.

1.x

dy

dx

= 3 −y

[

y= 3 +

c

x

]

2.

dy

dx

=x(1− 2 y)

[

y=^12 +ce−x

2 ]

3.t

dy

dt

− 5 t=−y

[

y=

5 t

2

+

c

t

]

4.x

(

dy

dx

+ 1

)

=x^3 − 2 y,givenx=1 when

y= 3

[

y=

x^3

5

−

x

3

+

47

15 x^2

]

5.

1

x

dy

dx

+y= 1

[

y= 1 +ce−x

(^2) / 2 ]
6.
dy
dx
+x= 2 y
[
y=
x
2

• 1
  4
  +ce^2 x
  ]
  48.4 Further worked problems on
  linear first order differential
  equations
  Problem 4. Solve the differential equation
  dy
  dθ
  =secθ+ytanθgiven the boundary condi-
  tionsy=1 whenθ=0.
  Using the procedure of Section 48.2:
  (i) Rearranging gives
  dy
  dθ
  −(tanθ)y=secθ, which
  is of the form
  dy
  dθ
  +Py=QwhereP=−tanθ
  andQ=secθ.
  (ii)
  ∫
  Pdx=
  ∫
  −tanθdθ=−ln(secθ)
  =ln(secθ)−^1 =ln (cosθ).
  (iii) Integrating factor e
  ∫
  Pdθ=eln(cosθ)=cosθ
  (from the definition of a logarithm).
  (iv) Substituting in equation (3) gives:
  ycosθ=
  ∫
  cosθ( secθ)dθ
  i.e. ycosθ=
  ∫
  dθ
  (v) Integrating gives: ycosθ=θ+c, which is
  the general solution. Whenθ=0,y=1, thus
  1 cos 0= 0 +c, from which,c=1.
  Hence the particular solution is:
  ycosθ=θ+ 1 ory=(θ+1) secθ
  Problem 5.
  (a) Find the general solution of the equation
  (x−2)
  dy
  dx


• 
  

  3(x−1)
  (x+1)
  y= 1
  (b) Given the boundary conditions thaty= 5
  whenx=−1, find the particular solution of
  the equation given in (a).
  (a) Using the procedure of Section 48.2:
  (i) Rearranging gives:
  dy
  dx

  
  


• 
  

  3(x−1)
  (x+1)(x−2)
  y=
  1
  (x−2)
  which is of the form
  dy
  dx
  +Py=Q, whereP=
  3(x−1)
  (x+1)(x−2)
  andQ=
  1
  (x−2)
  .
  (ii)
  ∫
  Pdx=
  ∫
  3(x−1)
  (x+1)(x−2)
  dx, which is
  integrated using partial fractions.
  Let
  3 x− 3
  (x+1)(x−2)
  ≡
  A
  (x+1)

  
  


• 
  

  B
  (x−2)
  ≡
  A(x−2)+B(x+1)
  (x+1)(x−2)
  from which, 3x− 3 =A(x−2)+B(x+1)