differential equation

452 DIFFERENTIAL EQUATIONS
Thus, the particular solution is: y x =−ln x + 2
or y =−x(ln x − 2) or y = x(2 − ln x)
Problem 2. Find the particular solution of the
equation: x dy
dx = x2 + y 2
, given the boundary
y
conditions that y = 4 when x = 1.
Using the procedure of section 47.2:
(i) Rearranging x dy
dx = x2 + y 2
gives:
y
dy
dx = x2 + y 2
which is homogeneous in x and
xy
y since each of the three terms on the right hand
side are of the same degree (i.e. degree 2).
(ii) Let y = vx then dy
dx = v + x dv
dx
(iii) Substituting for y and dy
dx
dy
dx = x2 + y 2
gives:
xy
v + x dv
dx = x2 + v 2 x 2
x(vx)
= x2 + v 2 x 2
vx 2
(iv) Separating the variables gives:
in the equation
= 1 + v2
v
x dv
dx = 1 + v2
− v = 1 + v2 − v 2
= 1 v
v v
Hence, v dv = 1 x dx
Integrating both sides gives:
∫ ∫ 1 v2
v dv = dx i.e.
x 2 = ln x + c
(v) Replacing v by y x gives: y 2
= ln x + c, which
2x2 is the general solution.
16
When x = 1, y = 4, thus: = ln 1 + c from
2
which, c = 8
Hence, the particular solution is: y2
2x 2 = ln x + 8
or y 2 = 2x 2 (8 + ln x)
Now try the following exercise.
Exercise 181 Further problems on homogeneous
first order differential equations
1. Find the general solution of: x 2 = y 2 dy
dx
[− 1 ( x 3
3 ln − y 3 ) ]
x 3 = ln x + c
2. Find the general solution of:
x − y + x dy
dx = 0 [y = x(c − ln x)]
3. Find the particular solution of the differential
equation: (x 2 + y 2 )dy = xy dx, given that
x = 1 when y = 1.
( [x 2 = 2y 2 ln y + 1 )]
2
4. Solve the differential equation: x + y
y − x = dy
dx
⎡ (
⎣ −1 2 ln 1 + 2y ) ⎤
x − y2
x 2 = ln x + c⎦
or x 2 + 2xy − y 2 = k
5. Find the particular ( solution ) of the differential
2y − x dy
equation:
= 1 given that y = 3
y + 2x dx
when x = 2. [x 2 + xy − y 2 = 1]
47.4 Further worked problems on
homogeneous first order
differential equations
Problem 3. Solve the equation:
7x(x − y)dy = 2(x 2 + 6xy − 5y 2 )dx
given that x = 1 when y = 0.
Using the procedure of section 47.2:
dy
(i) Rearranging gives:
dx = 2x2 + 12xy − 10y 2
7x 2 − 7xy
which is homogeneous in x and y since each of
the terms on the right hand side is of degree 2.
(ii) Let y = vx then dy
dx = v + x dv
dx