Linear Equation of the Second Order with Variable Coeffi cients 5-25
11.
2
2584
2
44(64)0.
dydy
xxxxy
dxdx
++++=
Ans:
4
8
3
sinlog4
2
x
ycexx
−
⎛⎞
=+
⎜⎟
⎝⎠
12.
2
2
2tan5sec.
x
dydy
xyxe
dxdx
−+=
Ans:
12
1
(sin6cos6)secsec
7
x
ycxcxxex=++
5.3 GENERAL SOLUTION OF
2
2
dydy
PQyR
dxdx
+=
+=
BY CHANGING THE INDEPENDENT VARIABLE
The differential equation is
2
2
dydy
PQyR
dxdx
++= (5.40)
Let the independent variable x be changed to z
where ().zzx=
We have
2
222
222
and
dydydzdydydzdydz
dxdzdxdxdzdxdzdx
⎛⎞
==+
⎜⎟
⎝⎠
Substituting in (5.40)
2
22
22
dzdydydzdydz
PQyR
dxdxdxdxdzdx
⎛⎞
+++=
⎜⎟
⎝⎠
2
22
22
dzdydzdzdy
PQyR
dxdzdxdxdz
⎛⎞
⎛⎞
+++=
⎜⎟
⎜⎟
⎝⎠
⎝⎠
Dividing by
2
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