Page 238 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Unit 15: Cauchy’s Problem and Characteristics for First Order Equations

Finally from (6) Notes
t
x = p + = 2 (e 1) +
t
or x = (2e 1) ...(15)
Substituting these values of x, y, p, q in the equation for z (t), we have

dz
t
t
t
t
= 2 (e 1) (2 e ) + (e + 1) (ve )
dt
dz
2
2
2t
or = 5 e 3 e t ...(16)
dt
on Integration of (16) we have
5 2
t
2
2t
z = (e 1) 3 (e 1) ...(17)
2
From (13) and (15)
t
t
x 2y = (2e 1) 2 (e 1)
or x 2y = , ...(18)
t
t
and y x = (e 1) (2e 1)
y x = e t
so using (18) we have by eliminating , we get
y x
t
e = ...(19)
2y x
t
Substituting these values of e and into equation (17) we have
2
5 2 y x 2 y x
z = x 2y 1 3 x 2y 1
2 2y x 2y x
5 2 5 2 2
y
= (y x ) (x 2 ) 3(y x )(x 2 ) 3(x 2 )
y
y
2 2
5 2 1 2 2
y
y
= (y x ) (x 2 ) 3(y x ) 3 (y x )
2 2
1 2 2 1 2 2 2
= (y 2yx x ) (x 4xy 4y ) 3y 3xy
2 2
3 2 1
y
= y 2xy y (4x 3 )
2 2
y
y
or z = (4x 3 ) ...(20)
2
is the solution of the equation (1).

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