Given y51+y−51=2x
⇒(51y−54−51y−56)dxdy=2
⇒y′(y51−y−51)=10y
⇒y′(2x2−1)=10y
⇒y′x2−1=5y
Differentiating again
y′′(2x2−1)+y′22x2−12x=10y′
⇒y′′(x2−1)+xy′=5x2−1(y′)
⇒y′′(x2−1)+xy′−25y=0
⇒λ=1,k=−25
If 2x=y51+y−51 and (x2−1)dx2d2y+λxdxdy+ky=0, then λ+k is equal to
Held on 9 Apr 2017 · Verified 6 Jul 2026.
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