对于问题1°,我们有:(1.220)为全微分方程的充要条积是 aM(a, y) aN(a, y) 实际上,若(1.220)为全微分方程,常线在函数(x,y)使得(1.2.21)成立 M,=N,从而由M,N作考虑可微来即得My=x 这说明(1.2.22)是(1.2.20)为全微分方程作必要条仍 反之,我们对a=M两边关于x积分(y作为参数)得a(x,y)如下 u(, y)=/M(, y)dz +p(y) (1.2.23) 必中φp(y)为一个待定作y作可微函数.为确定它,把(1.2.23)两边对y利偏 数,并利用=N,得p(D)厘满足作方程 0u0 y) ay ay N(,y) dy 亦即要利φ(y)满足 doly) 0 M(z, ydc (1.2.24) 显然只要上式右端与x无关,将必对y积分一次即得φ(y),代入(1.2.23)就 找到必全微分为(1.2.20)左端作u(x,y).取实上,由于M(x,y)考虑可微,因 此根据条仍(1.222)我们有 aN(, y) a az Ldy/M(a,y)dz) ON(x,y)o「o M(, y) x aN(, y aM(a, y) 从而将(1.2.24)两边对y积分得出φ(不意加意积分常数)项代入(1.2.23) 就得到(1.220)作通解 (a, y)dr +p(=c (1.2.25) 这不仅回答了问题29,而且外出了条仍(1.2.22)作充分来证明 完全类似地通解也可以表示为 u(, y)=/N(, y)dy+v(ar)=c (1.2.26)\GH1 ◦ , |}y (1.2.20) `|@ ∂M(x, y) ∂y = ∂N(x, y) ∂x . (1.2.22) EFf, @ (1.2.20) 'Ltuv3, :"# u(x, y) Ql (1.2.21) r, ^sy ∂u ∂x = M, ∂u ∂y = N, WU M, N noOt;sl My = uxy = Nx, (~ (1.2.22) (1.2.20) 'Ltuv3Xok. 2R, |} ∂u ∂x = M )hc\ x @u (y 'F#> l u(x, y) qw u(x, y) = Z M(x, y)dx + ϕ(y), (1.2.23) XI ϕ(y) '*j6 y Ot"#. 'N60< j (1.2.23) )h y Y/ s#< _Y{ ∂u ∂y = N< l ϕ(y) 78v3 ∂u ∂y = ∂ ∂y Z M(x, y)dx + dϕ(y) dy = N(x, y), ^soY ϕ(y) 78 dϕ(y) dy = N(x, y) − ∂ ∂y Z M(x, y)dx. (1.2.24) `gVofdg
N x Kc, bX y @usl ϕ(y), de (1.2.23) $ LMXLtu' (1.2.20) ￾
u(x, y). Ef, \ M(x, y) noOt, ' cmk (1.2.22) |}y ∂ ∂x · N(x, y) − ∂ ∂y Z M(x, y)dx ¸ = ∂N(x, y) ∂x − ∂ ∂x · ∂ ∂y Z M(x, y)dx ¸ = ∂N(x, y) ∂x − ∂ ∂y · ∂ ∂x Z M(x, y)dx ¸ = ∂N(x, y) ∂x − ∂M(x, y) ∂y ≡ 0. WUb (1.2.24) )h y @ulZ ϕ (tt@u#) de (1.2.23) $lM (1.2.20) ' u(x, y) = Z M(x, y)dx + ϕ(y) = c (1.2.25) (ChMGH 2 ◦ , U0ZMk (1.2.22) u; . L91' eOr' u(x, y) = Z N(x, y)dy + ψ(x) = c (1.2.26) 13