Page 336 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Unit 20: Higher Order Equations with Constant Coefficients and Monge’s Method

Notes
dq s dx dp s dy 4
sec y = 2q tan y
dy dx
4
4
or dq dx sec y dp dy 2 tan y dx dy = s (dx 2 sec y dy 2 )
q
Subsidiary equations are
4
dx 2 sec y dy 2 = 0 ...(1)
4
dq dx sin y dp dy 2 tan y dx dy = 0 ...(2)
q
From (1) x = tan y + . ...(3)
x = tan y + . ...(4)

From (2) and (3)
2
2
4
y
sec y dq dy sec y dp dy 2 tan sec y dy 2 = 0
q
2
q
or dq sec y dp 2 tany dy = 0
2
y
or cos y dq dp 2 sin cosy dy = 0
q
2
or q cos y p = C ( f x tan )
y
dx dy dz
= 2
y
1 cos y ( f x tan )
2
dx sec y dy dz
or =
y
2 ( f x tan )
1 2
( f x tan )(dx sec y dy ) = dz
y
2
y
( F x tan ) 2z = K.
y
y
or ( F x tan ) 2z = (x tan ) from (4)
The solution is
y
y
z = (x tan ) (x tan ).
1 2
Self Assessment

Solve the following differential equations by Monge’s method
2
2
17. 2x r 5xys 2y t 2(px qy ) 0
18. pt qs q 3

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