Question:
General solution of x log x (dy)/(dx) + y = (2)/(x) log x is
General solution of x log x (dy)/(dx) + y = (2)/(x) log x is
Options:
- y log x = -2/x (1 – log|x|) + C
- y log x = 2/x (1 – log|x|) + C
- y log x = 2/x (1 + log|x|) + C
- y log x = -2/x (1 + log|x|) + C
- y log x = -2/x (1 – log|x|) + C
- y log x = 2/x (1 – log|x|) + C
- y log x = 2/x (1 + log|x|) + C
- y log x = -2/x (1 + log|x|) + C
Correct Answer: d)
Explanation:
1 wlogz 2 +y= Llogx
‘ Di√iding by xlogx we get ‘
di
‘ rrr led i
it is a linear diff equation we get,
ods ods
1 yel ior = 2ef Trdz +c
ylogz = 2 logede +e
i)
H ylogz = 2—£ — 4 +c \
‘ ylogz = —2(1 + logz) + ¢
f 1 1 1 d
Since Zlogzdz = log da — J Zde—_logede
J Slogzde =loge — Lldz +c
‘ J Slogedz = Llogz —L +e
1 wlogz 2 +y= Llogx
‘ Di√iding by xlogx we get ‘
di
‘ rrr led i
it is a linear diff equation we get,
ods ods
1 yel ior = 2ef Trdz +c
ylogz = 2 logede +e
i)
H ylogz = 2—£ — 4 +c \
‘ ylogz = —2(1 + logz) + ¢
f 1 1 1 d
Since Zlogzdz = log da — J Zde—_logede
J Slogzde =loge — Lldz +c
‘ J Slogedz = Llogz —L +e