1.

If ydydx=x⎡⎢⎢⎢⎢⎢⎢⎣y2x2+ϕ(y2x2)ϕ′(y2x2)⎤⎥⎥⎥⎥⎥⎥⎦, x>0, ϕ>0 and y(1)=−1, then ϕ(y24) is equal to

Answer»

If ydydx=x⎡⎢
⎢
⎢
⎢
⎢
⎢
⎣
y2x2+ϕ(y2x2)ϕ′(y2x2)⎤⎥
⎥
⎥
⎥
⎥
⎥
⎦
,
x>0, ϕ>0 and y(1)=−1, then ϕ(y24) is equal to



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