452 DIFFERENTIAL EQUATIONSThus, the particular solution is:y
x=−lnx+ 2ory=−x(lnx−2)ory=x( 2 −lnx)Problem 2. Find the particular solution of theequation:xdy
dx=x^2 +y^2
y, given the boundaryconditions thaty=4 whenx=1.Using the procedure of section 47.2:(i) Rearrangingxdy
dx=x^2 +y^2
ygives:dy
dx=x^2 +y^2
xywhich is homogeneous inxandysince each of the three terms on the right hand
side are of the same degree (i.e. degree 2).(ii) Lety=vxthendy
dx=v+xdv
dx(iii) Substituting for y and
dy
dxin the equationdy
dx=x^2 +y^2
xygives:v+xdv
dx=x^2 +v^2 x^2
x(vx)=x^2 +v^2 x^2
vx^2=1 +v^2
v(iv) Separating the variables gives:xdv
dx=1 +v^2
v−v=1 +v^2 −v^2
v=1
vHence,vdv=1
xdxIntegrating both sides gives:
∫
vdv=∫
1
xdxi.e.v^2
2=lnx+c(v) Replacingvbyy
xgives:y^2
2 x^2=lnx+c, which
is the general solution.Whenx=1,y=4, thus:16
2=ln 1+cfrom
which,c= 8Hence, the particular solution is:y^2
2 x^2=lnx+ 8ory^2 =2x^2 ( 8 +ln x)Now try the following exercise.Exercise 181 Further problems on homoge-
neous first order differential equations- Find the general solution of: x^2 =y^2
dy[ dx
−1
3ln(
x^3 −y^3
x^3)
=lnx+c]- Find the general solution of:
x−y+xdy
dx=0[y=x(c−lnx)]- Find the particular solution of the differen-
tial equation: (x^2 +y^2 )dy=xydx, given that
x=1 wheny=1.
[
x^2 = 2 y^2
(
lny+1
2)]- Solve the differential equation:
x+y
y−x=dy
dx
⎡⎣−1
2ln(
1 +2 y
x−y^2
x^2)
=lnx+cor x^2 + 2 xy−y^2 =k⎤⎦- Find the particular solution of the differential
equation:(
2 y−x
y+ 2 x)
dy
dx=1 given thaty= 3whenx=2. [x^2 +xy−y^2 =1]47.4 Further worked problems on
homogeneous first order
differential equationsProblem 3. Solve the equation:
7 x(x−y)dy=2(x^2 + 6 xy− 5 y^2 )dx
given thatx=1 wheny=0.Using the procedure of section 47.2:(i) Rearranging gives:dy
dx=2 x^2 + 12 xy− 10 y^2
7 x^2 − 7 xy
which is homogeneous inxandysince each of
the terms on the right hand side is of degree 2.(ii) Lety=vxthendy
dx=v+xdv
dx