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Question Number 71516 by oyemi kemewari last updated on 16/Oct/19
5y″=(1+y′2)32pleasesolvethediffertialequation
Answered by mind is power last updated on 17/Oct/19
letz=y′5z′=(1+z2)32⇒dz(1+z2)32=dxz=tg(u)dz=(1+tg2(u))du⇒du1+tg2(u)=15dx⇒cos(u)=x5+vu=arcos(x5+v)z=tg(u)=1−(x5+v)2x5+vy=∫zdx=∫1−(x5+v)x5+vdx=5∫1−uudu,1−u=r⇒du=−2rdr5∫−2r21−r2dr=−10∫−1+11−r2dr=10r−10∫(12(1−r)+12(1+r))dr=10r−5ln∣1−r2∣y=101−(x5+v)−5ln∣x5+v∣+c
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