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Find the order of the D.E (y’’)^2+25y’=0...

Find the order of the D.E `(y’’)^2+25y’=0`.
a) 1
b) 2
c) 4
d) 3

Text Solution

Verified by Experts

The correct Answer is:
A, B, C, D
a. Since `mv=(M+m)v'`
`rarr v'=(mv)/(M+m)`

b. Velocity of the particle w.r.t C before the collision `=v-v'`
`=v-(mv)/(M+m)=(Mv)/(M+v)`
`c. Velocity of the rod w.r.t C before collision
`=-(Mv)/(M+m)`
`d. :. x_(cm)=(m_1x_1+m_2x_2)/(m_1+m_2)`
`=(ML)/(2(M+m))`

`:.` Angular momentum of body
`=mvr`
`=m(Mv)/(M+m).(ML)/(2(M+m))`
=`(M^2mvL)/(2(M+m)^2)`
`:.` Angular momentum of rod about cm.
`=M.(mv)/(M+m).1/2 m/(M+m)
`=(Mm^2vL)/(2(M+m)^2)`
`[:.r=1/2-(ML)/(2(M+m))=1/2 m/(M+m)]`
e. M.I. =`I_1+I_2`
`=-m[(ML)/(2(M+M))]^2+(ML^2)/12+M[(ML)/(2(m+M))]^2`
`=(mM^2L^2)/(4(m+M(]^2+(ML^2)/12+(Mm^2L^2)/(4(M+m)^)`
=(3mM^2L^2+M(m+M)^2L^2+3Mm^2L^2)/(12(M+m))`
`=(M(M+4m)L^2)/(12(M+m))`
f. we hav
`V_(cm)=(mv)/(M+m)`
`:. Iw=mvr=m.v. (ML)/(2(M+m))`
`:.w=(MvML)/(2(M+m)).(12(M+m))/(M(M+4m)L^2`
`=(6mv)/((M+4m)L)`
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