Question

高数求解！

Answer 1

。

Answer 2

dy/dx -y/(2x) =[1/(2y)].tan(y^2/x)
let
u= y^2/x
xu = y^2
x.du/dx +u = 2y.dy/dx
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dy/dx -y/(2x) =[1/(2y)].tan(y^2/x)
2y.dy/dx -y^2/x =tan(y^2/x)
x.du/dx +u - u = tanu
x.du/dx=tanu
∫ du/tanu = ∫ dx/x
ln|sinu| = ln|x| + C
y|x=2 = √(π/2) => u|x=2 = π/4
ln|sin(π/4)| = ln2 + C
-(1/2)ln2 =ln2 +C
C= -(5/2)ln2
ie
ln|sinu| = ln|x| -(5/2)ln2
sinu = x/2^(5/2)
sin(y^2/x) =x/2^(5/2)
y^2 = x.arcsin[x/2^(5/2) ]