Eliminating p from this and the given relation. We get
()
2
2422
–4yxpxp=
()
2
2222
–4
⎛⎞
=
⎜⎟
⎝⎠
c
yxcxx
x
()
2
2
–4yccx=
The general solution is
()
2
2
–4yccx=
Example 4.2.9 Solve y+px=p
2
x
4
.
Solution y+px=p
2
x
4
y=−px+p
2
x
4
Differentiating w.r.t. x and denoting
dy
dx
by p we get
423
–24
dpdp
ppxpxxp
dxdx
=−++
()
233
2–1–20
dp
ppxxxp
dx
+=
()()
33
21–21–20
dp
pxpxxp
dx
+=
Ch04.indd 16Ch04.indd 1612/9/2011 11:49:03 AM12/9/2011 11:49:03 AM
Differential Equations of the First Order but not ...
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