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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