Differential Equations of First Order and First Degree 2-61
Put
1
2
1
tan,
1
ytdydt
y
−
==
+
()
1
1
2
2
2tan
tan
1
2
1
.
2
t
t
y
y
cedt
ce
xece
−
−
=+
=+
=+
∫
Example 2.1.69 Solve
2
costan.
dy
xyx
dx
+=
[JNTU 1999S]
Solution Writing the equation as
22
sectansec
dy
xyxx
dx
+=
Here
22
()sec,()tansec.PxxQxxx==
This is linear in y. Also,
2
tan
()sectan
Integrating factor
x
Pxdxxdxx
e
==
=
∫∫
The general solution is
tan2tan
tansec
xx
yecxxedx=+
∫
Put
2
tan,secxtxdxdt==
(1)
t
t
ctedt
cte
=+
=+−
∫
Integrating by parts
tan
(tan1).
x
ycex
−
=+−
Example 2.1.70 Solve
40.xyy++=
′
[JNTU 2001]
Solution Writing the equation in the form
14
,
dy
y
dxxx
+=−
Ch02.indd 61Ch02.indd 6112/9/2011 ...
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