一阶线性微分方程求特解（附图）
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let
u= (x^3+1)y
du/dx = (x^3+1) dy/dx + 3x^2. y
//
y' +3x^2.y/(x^3+1) = y^2.(x^3+1). sinx
(x^3+1)y' +3x^2.y = y^2.(x^3+1)^2. sinx
du/dx = u^2 .sinx
∫ du/u^2 = ∫ sinx dx
1/u = cosx +C
1/[(x^3+1)y] = cosx +C
y(0) =1
1= 1 +C
=> C=0
1/[(x^3+1)y] = cosx
y= 1/[cosx .(x^3+1)]
u= (x^3+1)y
du/dx = (x^3+1) dy/dx + 3x^2. y
//
y' +3x^2.y/(x^3+1) = y^2.(x^3+1). sinx
(x^3+1)y' +3x^2.y = y^2.(x^3+1)^2. sinx
du/dx = u^2 .sinx
∫ du/u^2 = ∫ sinx dx
1/u = cosx +C
1/[(x^3+1)y] = cosx +C
y(0) =1
1= 1 +C
=> C=0
1/[(x^3+1)y] = cosx
y= 1/[cosx .(x^3+1)]
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y（0）＝1还要求特解 答案是y＝secx/（x^3＋1）。。。
追答
let
u= (x^3+1)y
du/dx = (x^3+1) dy/dx + 3x^2. y
//
y' +3x^2.y/(x^3+1) = y^2.(x^3+1). sinx
(x^3+1)y' +3x^2.y = y^2.(x^3+1)^2. sinx
du/dx = u^2 .sinx
∫ du/u^2 = ∫ sinx dx
1/u = cosx +C
1/[(x^3+1)y] = cosx +C
y(0) =1
1= 1 +C
=> C=0
1/[(x^3+1)y] = cosx
y= 1/[cosx .(x^3+1)]
= secx/(x^3+1)
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