Question

高数微积分求解

第七题，求解，谢谢

Answer 1

dy/dx=-1+sin(x+y)/x
设x+y=u，y=u-x，
dy/dx=du/dx -1，代入原方程得
du/dx=sinu /x
du/sinu =dx/x
du/[2sin(u/2)cos(u/2)] =dx/x
d(u/2) /[tan(u/2)cos²(u/2)] =dx/x
d[tan(u/2)] / tan(u/2) =dx/x
ln|tan(u/2)| =ln|x|+ln|c|
tan(u/2) =cx，u/2=arctan(cx)
x+y=2arctan(cx)
将x=π/2，y=0代入，得c=2/π
故x+y=2arctan(2x/π)

Answer 2

令 x+y = u, 则 y = u-x, dy/dx = du/dx-1
微分方程化为 x(du/dx-1) + x + sinu = 0, 即 xdu/dx = -sinu
du/sinu = -dx/x, ln(cscu-cotu) = -lnx + lnC
cscu - cotu = C/x, csc(x+y) - cot(x+y) = C/x
初值 y(π/2) = 0 代入， 1 = 2C/π, 得 C = π/2
则 csc(x+y) - cot(x+y) = π/(2x)

Answer 3

y = ∫[0,x] t f(x² - t²) dt
令u = x² - t²,du = -2t dt
当t = 0,u = x²；当t = x,u = 0
y = ∫[x²,0] t f(u) * du/(-2t)
= 1/2 ∫[0,x²] f(u) du
dy/dx = 1/2 [2x * f(x²) - 0]
= x f(x²)
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另一种方法
令z² = x² - t²,2z dz = -2t dt => dt = -z/t dz
当t = 0,z = x；当t = x,z = 0
∫[0,x] t f(x² - t²) dt
= ∫[x,0] t f(z²) * (-z/t) dz
= ∫[x,0] -z f(z²) dz
= ∫[0,x] z f(z²) dz
dy/dx = x f(x²) - 0
= x f(x²)