(87)-高数专项练题7_08.2026考研数学高途王喆全程班_赠送2025课程_25考研数学（三）全年智达班_{2}--资料

2024 专项刷题班第四章 常微分方程x  y x  y 例4.1 微分方程 y  sin  sin 的通解是 . 2 2 | =s - m 9 sm xty smt sm β= 2 10s Ʃ+ , β smπ - B ' - - 2 州 xy xY xey x = 8 +Ʃ Ʃ ⼀ 2 vs sm 2 z = g . ( osƩ δ mi = ±| dx sl ) == →>Ʃ - 2 冲Ʃ →/ )d ) . d 三 smil ( t ) Ʃ (cid:15482) |u |(s l - 10 = -2sm + C      例4.2 满足 f  x  xf   x  x的函数 f x ________. x t t - λ = - = , → fi - t) - t - fit ) = - t (cid:15482) f' ix ) - xfin = - x →fi - x ) = xfix 1 - x 代⼊原⽅程 fix [xfix x (cid:15482) x fix xi (cid:15482) 1 + x 1 - ) = x (+ ) ) = x+ x fix x+ (cid:15482) 1 = 2 1Hx ax 时 f =| 品 → * ax = Ʃ|nG x = + ) + x - arltmx + c  例4.3 求微分方程 ydx  x  x2  y2 dy  0, (y  0)的通解 y 解 → : Ydx = lo t xreyp ) ay x =xt xeys xtxep = ⼆年 P → ++ y d (u y) 令 年 y - . y 代⼊ U y uty = , , = * Y . ut → u = ts 1 y = s f(cid:15482) ieuu f yy h ut ieur= lulMl + lhC= licly (cid:15482) u+ (tu = =± CY = C . φ (C . =± C)U+ 1+ W = G . y = + cy P 可 y xty== G (cid:15482) x+ . = 平⽅ → xty= = C . - x , 4 2 yP , Pye : > x * = -2 C x+ x c . ) (cid:15482) 1 CP 2Gx = _ (cid:15482) 9 1 + 2 C x . = GE          例4.4  x , x 为已知的连续函数，则方程 y  x y  x  x 通解是_____． Seix ) dx 1 uindx cfe ①(+) .ω ix 1x tc ) φ= e , -e e " . fee ) c * e*do tc = Jtdt letedt ett . ettc - = = t t - λ ↓ ( t ) C = t -1 + eel, 4 ( een! - φ(X) : - (( ew 1 ) + c ) =φ x1-1 + ce例4.5 求方程 xy(4)  y(3)  0的通解. 3) = pi 令 φ P 则Y 性 → x . P = 0 = p - . 1 iax / P ⽚ P → P c o dx tc ) → - =0 = e , lmxl → P e C C (λ =± Cx Cx = - = = " 叫 Gx → = = x " =| →φ c. xx + Cz → Y = / ( Ʃ x + C 2 ) dx =Ʃ . x + 2 x + G →φ=|③ x =☆ 4 +Ʃ= + Crx + ( 3 ) dx x x + ( 3 x + C 41 例4.6 已知 y  cos2x  xcos2x是二阶常系数非齐次微分方程一个解， 4 则该方程是( B ) (A) y  4 y  sin2x (B) y  4 y  sin2x 1 1 (C) y  4 y  sin2x (D) y  4 y  sin2x 4 4 E 代⼊ 1 1 (cid:49455)1 ①) φ= 10s2x . 法 _ : 1032× - ③ . * ⻬ Gerxtaerx ⾮⻬通 C 4 C 叫ztY : ① r .tr 2 φ= 法 = : = . . + 2 ⻬通中⽆⼀式 形式 ② r = rzzr acoszx . ∵ erxexrotG * φ= c ⼀书 是 y xc 0 s2x , ∴ 是⼀个⻬次解 ③ r 1 =α+βi r 2 = 5 - βi ∴ (os2x q = C et* C 0s β xtrexsm βx , .各通 之 r 2i Gcos 2x 2 sm2x r 2 2 = - + ∴ = . 特征⽅程午 : 4 = 0 =ψ ⻬次 +4 " = 叫 " ⼀ 原公程 ψ 4y f 代⼊ Y 1oszx : + = ⼭ * , f Sm2x (cid:15482) 1 × =  例4.7 设函数 y  y x 满足微分方程 y  4 y  4 y  0和初始条件  y  0   2, y  0   4，求广义积分 y  x  dx. 0 ' γ+ 特征⽅程 : 4 r+4 = 0 (cid:15482) ( r+2) =0 → r . =γ . = - 2 2x 2X - . - Cax Yun + e e . = C qi : φ 0) = 2 , 0 ) = -4 1 . C 1 = 2 C 2 = 2X φ ⼀ . *)= . e 2 |^∞ ex 9t e -^ : 1 [ q .* dx = 2 . dx = - (cid:49478) ∞ x 2 d ) (cid:49478) ⼼ 0 =* ( t" - 1 e = = - (cid:49478)例4.8 求微分方程 y  4 y  sin3xsin x  cos2 x的通解. γ= 特征⽅程 : - 4 =0 → 1 . 22 , γ 2 =- 2 : φ= C e 2x + 2e 2x sm 2 . sm β= - Ʃ [ cos(σ+β ) -10 (σ - β ] . Ʃ \+ Cos2X sm 3x.smX +c 0 x = - [os 4x - cos2x ] + 2 三 Ʃ ( 0s 4x +co> 2X + = - " 9 4Y Ʃ cos 4x ① . - = - * * Fi Hi = acos4 xtbsm4x , M , 代⼊ ① → a = b , = 0 1 . , * Ficos : 9 4x = i" × y -4 Y = ( 0s2 - 9 Ʃ φ % * = 210s2 x + b =sm2 X * 代⼊ ②(cid:15482) G 2 = b →= a ⼀⽚ 上 φ* los2x = , " ψ 44 Ʃ ③ - = * i ; Ʃ : φ = A (cid:15482) A = - ∴ φ * = - ; - - : * = φ . * + φ * +φ ; * = icosφ x Ʃlos2 x : 通解 : φ= G + e 2 x + ( os 4 x-δ cos 2x- Ʃ ,例4.9 设函数 f (x), g(x)满足 f (x)  g(x), g(x)  2ex  f (x)，且  g(x) f (x)   f (0)  0, g(0)  2，求  dx.   0 1  x (1  x)2   . -. a =| qu xlvix 保 ux xrdx f ↑ ↓ f^ t x f itaixn kax - xi f iit " . . - . fi feo π ) ) = - (以 fix 1 = 94) (cid:15482) qix ) = f" ) , ⽽ qix 1 = 2e * - fi +) , " f ex fi 1 (x) = 2 - *) .f" fex * : (*+ ) = 2 . γ=+ 特征 = 1 = 0 (cid:15482) γ= i r 2 =- i T C C smx : = cosx + 2 . * fY fx × * y = ae ' 代⼊ 1 + 1 = 2 e 中 (cid:15482) a = 1 . ex 性 . fi ∴ *1 = C , cosxt ( 2 Smx t ex ⼜ fo 1 = 0 , q . 0) = fi 0 )= 2 ∴ . C . = - 1 ( 2= 1 f f ex ltex cosxtsmx ( ) ∴ ×|= - + π = = 原 ∴例4.10 求微分方程 y 6  2 y 3  y  0的通解. 特征⽅程 γ 6 + 2γ 3 + 1 = 0 → ( r 3 + 1 ] = 0 1 ± 1 4 l± 给之 P ) ≥ - [ 1 ( r = -r+ 1 == r r + 1 = 0 r = = → 2 2 1+ o 3i 1 ⼀场之 (cid:15482) V = - | V 2 =- | r 3 =γ 4 = γ 5 =γ 0 = . 2 2 起 esmBx cos * GxeX q = C e + + ( 3 . ③ x + 4 . 10 eismiix xt t ( xe Ʃ ③ ( 0 5 s  例4.11 设 f (u)有二阶连续导数,z  f x2  y2 满足 2 2  z  z   x2  y2 ，求 2 2 x y z x 的表达式. f z = u ( xtyl - y f 令 U= x 2 y 2 _ u x az fiu ) = . x 2 2 中 xty 北 xey - X = " 以 efiu = 2)0 ep ↑ ) f 2 eyz x 2 " tfiu 时 f ul ) = ( . 3 =+ } (x y 收 y xx = "( f' f u) , t ial . 3 2 X yp ( x =+ y ]xeyz 代⼊ ( 'typ f" fiul x (cid:15482) ul = ( + ⿏ 2 f" fl π × 缺型 , 5 ult . = , + 令 P = f ' u ) f" (u) = 1 E = E ( x + y =|P C . 1 n xeyz + Cz , 2 aP p u → + = letau ' efaau u c ut al dutcd → P . . + = = ' aG fiul u → + = fll 1 ( a π 4 ( lm 1 G → = + ) u = u + , +例4.12 （仅数一数二）已知 y (x)  ex , y (x)  u(x)ex 是二阶微分方程 1 2 (2x  1) y  (2x  1) y  2 y  0的解，若u(1)  e,u(0)  1，求 u ( x ) ， 并写出该微分方程的通解. Ur = uxiet 4 i = Wix ) - et + ux1 . eX W = N +ux1 上 ex , φ " = ω× 1e* + 2ω ix 1. e * + u1x1 . e * ) = ω "(x +2 Wix ) tuix)] ex i 代⼊原公程 ψ eX ' e e. (cid:15482) ( 2x - 1 ) [ "(x) +2 Wix ) + U(x) ] . - [2x+1) [ω (x) + U1×1 ] . * 2φ(x ) * . Wix + (cid:15482) ( 2x- 1 ) U"x ) (x - 3 ) ) =0 p " 令 ix P=ω ) =ω (x) .. (cid:15482) ( 2x-1 ) P + (2x -3 ) . P = ☆ IP → (2x +) . = - - 2x 3 P 2x- →|µ l dx = - 2× - e* (cid:15482) P C (2x -1 . = . ) ex : uix ) = [ . ( 2x-1 ) . (cid:49456) txdx -XeC ∴ u*1 = C . (2x - 1 ) . = - C . ( 2x+1 ) , e 2 ⼜ UC1|= e φ ( 0) = - , : . C . = ( C 2 = ex ∴ U(N) = - ( 2x+1) . ∴ φ 2 = - ( 2x+1 ) 原通解 φ= c ex C [ 2xt1 ) ∴ . - 2例4.13 设函数 f (x)在[1,)上连续.若由曲线 y  f (x)，直线 x  1, x  t(t  1)与 x 轴所围成的平面图形绕 x 轴旋转一周所成的旋转体  体积为V  t   t2 f  t   f  1  ，试求 y  f  x  所满足的微分方程，   3 2 φ=和 并求该微分方程满足条件 y  的解. x2 9 ⼀ ? ,ta f i ③ [ ife * f ↓ ) 对球导 ! D ut= ax - f = , > ③ xfii 达 fit tfit] 七 2 ) + ) | (cid:15482) = ( tfe f tfit =(cid:15482) 3 t =2 1+ ) , == ×φ+ xx (cid:15482) 3 2= dy + . → 3 2 x φ x 2 dx dl 3 _ 作恐 i xy (cid:15482) 3 2 = = dx y dy a 令 U = 元 =U + , dx ax x . 2 u (cid:15482) u+ x . = 3 = - 2 U → x = 3u - 3U = 3 u (u- 1) = xaxx fucu 3 _yu f ui ildu = 31 ux1 hc s + ( - β u 出 ucx → =β u 出 = → ncx ⼗ " 3 a- (cid:15482) =± ( x = C x . . u 9 1 - 3 九 (cid:15482) 4 x C x - 43 = . → C = . Ψ y 可 Y| ⼜ x=2 = x x : φ= C =+ - : = . 單 = 3 ltx