1 12 2 1 2 1 1 2 2 1 2 12习题六1. 指出下列各微分方程的阶数：(1)一阶 (2)二阶 (3)三阶 (4)一阶2. 指出下列各题中的函数是否为所给微分方程的解：(1)xy ' = 2 y , y = 5x 2 ;解：由 y = 5x 2得 y ' = 10x 代入方程得x ⋅10x = 2 ⋅ 5x 2 = 10x 2故是方程的解.(2) y ' + y = 0, y = 3sin x - 4 cos x ;解： y ' = 3cos x + 4 s in x ;y ' = -3sin x + 4 cos x 代入方程得故是方程的解.-3sin x + 4 cos x + 3sin x - 4 cos x = 0 . (3) y ' - 2 y ' + y = 0,y = x 2e x ;解： y ' = 2x e x + x 2e x = (2x + x 2 )e x,代入方程得2e x ≠ 0 . 故不是方程的解.(4) y ' - (+ ) y ' +y = 0,y ' = (2 + 4x + x 2 )e xy = C e 1x + C e 2 x .1 2 1 21 2 y ' = C e 1x + C e 2 x ,y ' = C 2e x 1+ C 2e 2 x解：1 12 21 12 2代入方程得C 2e 1x + C 2e 2 x - (+ )(C e 1x + C e 2 x ) + (C e 1x + C e 2 x) = 0. 故是方程的解.3. 在下列各题中，验证所给二元方程为所给微分方程的解：(1)(x - 2 y ) y ' = 2x - y , x 2 - xy + y 2 = C ;证：方程x 2 - xy + y 2 = C 两端对 x 求导： 2x - y - xy ' + 2 yy ' = 0 y ' =2x - y 得x - 2 y 代入微分方程，等式恒成立.故是微分方程的解.(2)(xy - x ) y ' + xy '2 + yy ' - 2 y ' = 0, y = ln(xy ). 证：方程 y = ln(xy ) 两端对 x 求导： y ' = 1 + 1y 'x y(*)y ' =得y x ( y -1) .(*)式两端对 x 再求导得x =0 ⎰ ⎰ y ' = - y ⎡ 1 +1 ⎤y -1 ⎢⎣ x 2 x 2 ( y -1)2 ⎥⎦将 y ', y ' 代入到微分方程，等式恒成立，故是微分方程的解.4. 从下列各题中的曲线族里，找出满足所给的初始条件的曲线：(1)x 2 - y 2 = C ,y = 5;解：当 x = 0 时，y =5.故 C =-25故所求曲线为： y 2- x 2= 25(2) y = (C + C x )e 2x ,y = 0, y '= 1.12x =0y ' = (C + 2C + 2C x )e 2xx =0解：212当 x =0 时，y =0 故有C 1 = 0 . 又当 x =0 时， y ' = 1.故有C 2 = 1. 故所求曲线为： y = x e 2 x. 5. 求下列各微分方程的通解：(1)xy ' - y ln y = 0 ; d y = 1 d x解：分离变量,得积分得y ln y x 1 d ln y = 1d xln y x得(2) y ' ln ln y = ln x + ln c ln y = cxy = e cx .解：分离变量，得积分得⎰ d y =⎰ d x得通解：-= -+ c .(3)(e x + y - e x )d x + (e x + y + e y )d y = 0 ;解：分离变量，得 e y1- e yd y =e y 1+ e x d x 积分得 -ln(e y -1) = ln(e x +1) - ln c 得通解为(e x +1)(e y -1) = c .(4) cos x sin y d x + sin x cos y d y = 0 ;cos x d x + cos y d y = 0解：分离变量，得 sin x sin y积分得ln sin y + ln sin x = ln cy = 0 得通解为sin y ⋅sin x = c .(5) y ' = xy ;解：分离变量，得 d y= x d x yln y = 1x 2 + c积分得得通解为(6)2x +1+ y ' = 0 ; 2 1 x 2 y = c e21(c = e c 1 )解： y ' = -2x -1积 分 得 得通解为y = ⎰ (-2x -1)d xy = -x 2 - x + c . (7)4x 3 + 2x - 3y 2 y ' = 0 ;解：分离变量，得积分得即为通解.(8) y ' = e x + y .3y 2d y = (4x 3 + 2x )d x y 3 = x 4 + x 2 + c解：分离变量，得e - y d y = e x d x⎰ e - yd y = ⎰e x d x积分得得通解为：-e - y= e x+ c . 6. 求下列各微分方程满足所给初始条件的特解：(1) y ' = e 2x - y ,解：分离变量，得x =0 ;e y d y = e 2x d x e y = 1e 2 x + c积分得以 x = 0, y = 0 代入上式得 2 . c = 1 2故方程特解为 e y = 1 (e 2 x +1)2 .(2) y 'sin x = y l n y , y x = π = e2.解：分离变量，得 d y y ln y c ⋅tan x=d x sin x 积分得x = π, y = ey = e2将 2 代入上式得c = 1tan x故所求特解为y = e2.= ⎝ ⎭ 7. 求下列齐次方程的通解：(1) xy ' - y -d y = y + 解：dx x = 0 ;u = ⇒ = u + x du令原方程变为x d x d ud x = d x x两端积分得ln(u u = ln x + ln c = cxy +x 即通解为：y += cx = cx 2(2) x d y = y ln yd x x ; d y = y ln y 解： d x x x u = yd y = u + xd u 令 x ， 则d x d ud x =d x 原方程变为u (ln u -1) x 积分得ln(ln u -1) = ln x + ln cln u -1 = cx ln y-1 = cx x 即方程通解为y = x e cx +1(3)(x 2 + y 2 )d x - xy d x = 0⎛ y ⎫2d y x 2 + y 2 1+ x ⎪解：u = y d x xy y x d y = u + xd u 令x ， 则 d x u + x d u d x = 1+ u 2原方程变为d x u x d u = 1 ,u d u = d x 即d x u x= + 1 x 1u 2 = ln x + ln c积分得2 y 2 x22 ln x 12 ln c 1故方程通解为y 2 = x 2 ln(cx 2 )(c = c 2 )(4)(x 3 + y 3 )d x - 3xy 2d y = 0 ;⎛y ⎫3 d y x 3 + y 3 1+ x⎪ = = ⎝ ⎭ d x 3xy 2 解：⎛ y ⎫23 ⎪ ⎝ ⎭ u = y d y = u + xd u 令x , 则 d x u +d u d x 1+ u 3x = 原方程变为d x 3u 2 3u 2 d x即1- 2u 3d u = x - 1ln(2u 3 -1) = ln x + ln c积分得 2 1y以 x 代替 u ，并整理得方程通解为 (5) d y =x + y d x x - y ;1+ y 2y 3 - x 3 = cx . d y =d x 解：x 1- yx u = y d y = u + x d u 令x ， 则 d x d x u + x d u =1+ u 原方程变为d x 1- u 分离变量,得1- u 1+ u 2 d u = 1 d xx arctan u - 1ln(1+ u 2 ) = ln x + ln c积分得2 yx 1x 2 + y 2= c e2arctany1 x. (c =)c 2以 代替 u ，并整理得方程通解为到1(6) y 'c ⎪ y = 1 x yd y =d x 解：d x = x +即d y y x = v x = yv , = v+ y d v令y ， 则d yd y , 原方程可变为v+ y d v= v d y y d v =即d y=d y 分离变量，得积分得即ln(v v⎛ y y= ln y - ln c .=y c⎫2- v ⎪ ⎝ ⎭ = v 2 +1 y 2 - 2 yv =yv = xc 2c 1 y 2 = 2c ⎛ x + c⎫ 以 代入上式，得⎝ 2 ⎭ 即方程通解为y 2 = 2cx + c 2. 8. 求下列各齐次方程满足所给初始条件的解：(1)( y 2 - 3x 2 )d y + 2xy d x = 0,2yx =0 ;d y = -d x解：x ⎛ y ⎫2 ⎪ ⎝ ⎭ u + x d u = - 2u令 y = ux ，则得 d x u 2 - 3u 2 - 3 d x分离变量，得u - u 3d u = x 积分得 -3ln u + ln(u -1) + ln(u +1) = ln cx u 2 -1即ln u 3 x= ln c - 3c 2 c 3 ⎰ 2 ⎰ 2 3 3 得方程通解为y 2 - x 2 = cy 3 以 x =0，y =1 代入上式得 c =1.故所求特解为y 2 - x 2 = y 3 . (2) y ' = x + y,y xy x =1 = 2.d y = u + xd u 解：设y = ux ， 则 d x d x u d u = d x原方程可变为积分得x 1 u 2 = ln x + ln c 2 . 得方程通解为y 2 = 2x 2(ln x + ln c ) 以 x =1，y =2 代入上式得 c =e 2.故所求特解为y 2 = 2x 2(ln x + 2) . 9. 利用适当的变换化下列方程为齐次方程，并求出通解：(1)(2x - 5 y + 3)d x - (2x + 4 y - 6)d y = 0 解：设 x = X +1, y = Y +1 ，则原方程化为2 - 5 Yd Y =2 X - 5Y d X 2 X + 4Y = X2 + 4 YXu = Y 令X ⇒ u + X d ud X = 2 - 5u2 + 4u ⇒ - 4u + 2 d u = d X4u 2+ 7u - 2 X⇒ ln X = - 1 (8u + 7) - 3 d u2 4u + 7u - 2 = - 1 ln(4u 2 + 7u - 2) +3 d u2 2 4u + 7u - 2 = - 1 ln(4u 2 + 7u - 2) + 1 ⎛ - 1 + 4 ⎫d u2 6 ⎰ u + 2 4u -1⎪⎝ ⎭= - 1 ln(4u 2 + 7u - 2) - 1 ln 4u -1 + ln c2 6 u + 2 1⇒ 6 ln X + 3ln(4u 2 + 7u - 2) + ln 4u -1= ln c (c = c 6 )⇒ X 6 (4u 2 + 7u - 2)3 ⋅ 4u -1= cu + 22 2 1u + 2 2⇒ X 6 (4u -1)4 (u + 2)2 = c代回并整理得⇒ X 3 (4u -1)2 (u + 2) = c ,(c = )(4 y - x - 3)2 ( y + 2x - 3) = c , (c = ) .2(2)(x - y -1)d x + (4 y + x -1)d y = 0;d y = - 解：d x x - y -1 4 y + x -1 作变量替换，令 x = X +1, y = Y + 0 = Y1- Yd Y = - X - Y = - X原方程化为令Y = uX ,则得d X X + 4Yu + Xd ud X1+ 4 YX= - 1- u ⇒ X d u 1+ 4u d X= - 1+ 4u 21+ 4u分离变量，得积分得-1+ 4u 1+ 4u 2 d u = d X x11 d(1+ 4u2 )ln X = -⎰1+ 4u 2 d u - 2 ⎰1+ 4u 2 = 1 arctan 2u - 1ln(1+ 4u 2 ) + c 2 2即2 ln X + ln(1+ 4u 2 ) + arctan 2u = c⇒ ln X 2 (1+ 4u 2 ) + arctan 2u = c代回并整理得ln[4 y 2 + (x -1)2 ] + arctan 2 y x -1 = c .(3)(x + y )d x + (3x + 3y - 4)d y = 0 ;d y = d v -1解：作变量替换v = x + y , 则d x d x 原方程化为d v-1 = - d x v 3v - 4 ⇒ d v =2(v - 2) d x 3v - 4 ⇒ 3v - 4 d v = d x 2(v - 2) ⇒ 3 ⎰ d v + ⎰ 1d v = ⎰ d x 2 v - 2 ⇒ 3v + ln(v - 2) = x + c21代回并整理得 (4) d y=⇒ 3v + 2 ln(v - 2) = 2x + c , (c = 2c 1)x + 3y + 2 ln(x + y - 2) = c . 1 +1 d x x - y .d u = 1- d y 解：令u = x - y , 则d x d x1 ⎢⎰ ⎦ ⎰⎡⎰ d x ⎣ ⎦⎣ ⎦原方程可化为分离变量，得 d u = - 1 d x u u d u = -d x 1 u 2= -x + c积分得21u 2 = -2x + 2c故原方程通解为1(x - y )2= -2x + c . (c = 2c )10. 求下列线性微分方程的通解：(1) y ' + y = e -x ;解：由通解公式y = e -⎰d x⎡e - x ⋅ e ⎰d x d x + c ⎤ = e -x ⎡⎰ e -x ⋅ e x d x + c ⎤ = e -x (x + c ) ⎢⎣⎰ (2)xy ' + y = x 2 + 3x + 2 ；⎦⎥⎣ ⎦解：方程可化为由通解公式得y ' + 1 y = x + 3 + 2 x x-⎰ 1d x ⎡2⎰ 1d x ⎤ y =e x⎢⎣⎰ (x + 3 + ) ⋅ e x x d x + c ⎥⎦ = 1 ⎡⎰ (x + 3 + 2) ⋅ x d x + c ⎤x ⎢⎣ x⎥⎦ = 1 x 2 + 3 x + 2 + c . 3 2 x(3) y ' + y cos x = e -sin x ;y = e -⎰cos x d x ⎡ 解：⎣ (4) y ' = 4xy + 4x ;e -sin x ⋅ e ⎰cos x d x d x + c ⎥⎤ = e -sin x (x + c ). y = e-⎰(-4 x )d x ⎡ 4x e ⎰(-4 x )d x d x + c ⎤ = e 2 x 2⎡⎰ 4x e -2 x 2d x + c ⎤ 解：⎢⎣⎰ ⎥⎦ ⎣ ⎦ = e 2x 2(-e -2x 2+ c )= c e 2x 2-1(5)(x - 2) y ' = y + 2(x - 2)3 ; d y - 1 y = 2(x - x )2解：方程可化为d x x - 2 y = e- -1d x x -2⎢⎣⎰ -1⎤ 2(x - 2)2e x -2d x + c ⎥⎦= e ln( x -2) ⎡⎰ 2(x - 2)2e -ln( x -2)d x + c ⎤ = (x - 2) ⎡⎰ 2(x - 2)d x + c ⎤ = (x - 2)3 + c (x - 2)(6)(x 2 +1) y ' + 2xy = 4x 2..x ⎣ ⎦ ⎰ ⎥ ⎪ ⎪ ⎢⎰ ⎣ ⎦ ' 2x4x 2解：方程可化为y +x 2 +1 y =x 2 +1 -⎰ 2 x d x ⎡ 2 2 x d x ⎤ y = e x 2+1 ⎢ 4x e ⎰ x 2 +1d x + c ⎥ ⎣⎰ x 2 +1 ⎦= -ln( x 2+1) ⎡⎰ 2⎤4x 3 + c 2e⎣ 4x d x + c ⎦ =3(x+1) 11. 求下列线性微分方程满足所给初始条件的特解： (1) d y + 1 y = 1sin x , d x x x y x =π = 1 ; -⎰ 1 d x ⎡ sin x ⎰ 1d x ⎤ 11 y = e x 解：⎢⎣⎰ e xd x + c ⎥⎦= ⎡⎰sin x d x + c ⎤ = [c - cos x ] x 以 x = π, y = 1代入上式得c = π -1，y = 1(π -1- cos x )故所求特解为 x .(2) y ' + 1 (2 - 3x 2) y = 1, x 3 y x =1 = 0. 2 - 3x 2解：d x = -x -2 - 3ln x + c- 2-3x 2 2∴ y = e ⎰ x 3 d x ⎡ ⎢⎰ e 2-3x d x x 3 d x + c ⎤ = e x -2 +3ln x ⎡⎰ e - x -2-3ln x d x + c ⎤⎣ ⎦⎣ ⎦ = e x -2 ⋅ x 3 ⎛ 1 e -x -2 + c ⎫ = x 3 ⎛ c e x -2 + 1 ⎫.⎝ 2 以 x =1,y =0 代入上式，得 c = - ⎭ ⎝ 2 ⎭1 2e . y = x 3 ⎛ 1 - 1 e x -2 ⎫ 2 2e ⎪ 故所求特解为⎝ ⎭ . 12. 求下列伯努利方程的通解：(1) y ' + y = y 2 (cos x - sin x );解：令 z = y 1-2 = y -1，则有d z + (1- 2)z = (1- 2)(cos x - sin x ) ⇒ d z- z = sin x - cos x d x z = e -⎰(-1)d x ⎡ ⎣d x(sin x - cos x )e ⎰(-1)d x d x + c ⎤⎥⇒ 1= c e x - sin x y即为原方程通解.= e x ⎡⎰ e-x (sin x - cos x )d x + c ⎤ = c e x - sin x (2) y ' + 1 y = 1(1- 2x ) y 43 3 .⎰ ⎦ x x 3⎢⎰ ⎦ 1 1 2⎰解：令z = y -3 ⇒d z- z = 2x -1 d x . z = e ⎰d x⎡ ⎣(2x -1)e -⎰d x d x + c ⎤⎥ = -2x -1+ c e x⇒ y 3 (c e x - 2x -1) = 1即为原方程通解.13. 求下列各微分方程的通解：(1) y ' = x + sin x ;解：方程两边连续积分两次得y ' = 1x 2 - cos x + c2 1y = 1x 3 - sin x + c x + c(2) y ' = x e x ;解：积分得⎰61 21y ' = y ' = ⎰ x e x d x = x e x - e x + c(x e x - e x + c )d x = x e x - 2e x+ c x + c y = (x e x - 2e x + c x + c )d x = (x - 3)e x- 1 c x 2 + c x + c⎰1 2(3) y ' = y ' + x ;2 12 3解：令 p = y ' ,则原方程变为p ' = p + x , p ' - p = x ,p = e ⎰d x⎡ -⎰d x⎤ = c e x - x -1 ⎣ x e d x + c 1 ⎦ 1y = (c e x - x -1)d x = c e x - 1x 2 - x + c故11 2 2 .(4) y ' = ( y ')3 + y ' ;y ' = p d p解：设y ' = p ， 则 d y原方程可化为 p d p = p 3 + p d y p ⎡d p - (1+ p 2 )⎤= 0 即⎢⎣ d y ⎥⎦ 由 p =0 知 y =c ，这是原方程的一个解.d p= 1+ p 2 ⇒ 当 p ≠ 0 时， d yd p 1+ p 2 = d y⇒ arctan p = y - c 1⇒ x = d ytan( y - c 1) = ln sin( y - c 1) - c 2'∴ y = arcsin(c e x ) + c (c = e c 2' )212(5) y ' = 1;x⎰x =1 x =1 ;(arcsin x + c )d x = x arcsin x c x + c . 1y ' = 1d x = ln x + c ' 解：⎰ x 1y = ⎰ (ln x + c 1')d x = x ln x - x + c 1'x + c 2 = x ln x + c 1 x + c 2 (c 1 = (-1+ c 1'))(6) y ' =; y ' = 解：⎰ x = arcsin x + c 1 1 2 (7)xy ' + y ' = 0 ;p ' + 1 p = 0 ⇒ d p + d x= 0解：令y ' = p ,则得 x p x ⇒ ln p + ln x = ln c 1p = c 1得 xy = c 1 d x = c ln x + c 故 ⎰ x 1 2 .(8) y 3 y ' -1 = 0 .y ' = p d p解：令 p = y ' ，则原方程可化为d y . y 3 p d p-1 = 0, p d p = y -3d y d y ⇒ 1 p 2 = - 1 y -2 + c1 ⇒ p2 = - y -2 + c2 2 2 1⇒ d y = d x ⇒ ±⎰ y d y = ⎰ d x⇒ ±= 2c 1 x + 2c 2⇒ = c x + c ⇒ c y 2 -1 = (c x + c )2.1211214. 求下列各微分方程满足所给初始条件的特解：(1) y 3 y ' +1 = 0, y = 1, y ' = 0 y ' = p d p解：令y ' = p ，则 d y , y 3 ⋅ p d p = -1 ⇒ p d p = - 1 d y原方程可化为d y y 3 ⇒ 1 p 2 = 1 y -2 + 1 c 2 2 2 1⇒ p 2 = y2 + c 11 y =x =1 x =1 ;由 x = 1, y = 1, y ' = p = 0 知， c 1 = -1，从而有' y = p =⇒y = ±d x由 x = 1, y = 1 ,得c 2 = 1 ⇒ = ± x + c 2故x 2 + y 2 = 2x 或 y =.(2)x 2 y ' + xy ' = 1, y = 0, y ' = 1 解：令 y ' = p ，则 y ' = p ' .p ' + 1 p = 1原方程可化为x x 2 -⎰ 1d x ⎡ 1 ⎰ 1 d x ⎤ 1p = e x ⎢⎣⎰ x 2 e d x + c 1 ⎥⎦ = x (ln x + c 1 )y ' = 1 (ln x + c )则x 1以 x = 1, y ' = 1代入上式得c 1 = 1y ' = 1(ln x +1)则x y = 1 ln 2x + ln x + c当 x =1 时，y =0 代入得c 2 = 0故所求特解为2 2y = 1ln 2 x + ln x 2 . (3) y ' = 1 , y x 2+1 x =0= y ' x =0 = 0 ;解： y ' = arctan x + c 1当 x = 0, y ' = 0 ，得c 1 = 0y =arctan x d x = x arctan x -xd x⎰⎰1+ x 2= x arctan x - 1ln(1+ x 2 ) + c22以 x =0,y =0 代入上式得c 2 = 0 故所求特解为 y = x arctan x - 1ln(1+ x 2 )2 .(4) y ' = y '2 +1, y x =0 = 1, y ' x =0 = 0 ;解：令 p = y ' ，则 p ' = y ' .xx =0 x =0 ;⎪x =0 x =0 .2原方程可化为p ' = p 2 +1 d pp 2 +1= d x arctan p = x + c 1 y ' = p = tan(x + c 1)以 x = 0, y ' = 0 代入上式得c 1 = k π .y = ⎰ tan(x + k π)d x = -ln cos(x + k π) + c 2以 x =0,y =1 代入上式得c 2 = 1故所求特解为y = -ln cos(x + k π) +1(5) y ' = e 2 y , y = y ' = 0 y ' = p d p解：令 y ' = p ，则原方程可化为即 d y . p d p= e 2 y d y p d p = e 2 y d y 1 p 2 = 1 e 2 y + 1 c 积分得2 2 2 1p 2 = e 2 y + c 以 x = 0, y = y ' = 0 代入上式得c 1 = -1, 则p = y ' = ± e 2 y-1= ±d xarcsin e - y = x + c c = π 以 x =0,y =0 代入得2 ，故所求特解为arcsin e - y = x + π2e - y = sin ⎛ π± x ⎫ = cos xy = ln sec x 即⎝ 2 (6) y ' = 3 ⎭ . 即 . y , y = 1, y ' = 2解：令y ' = p , y ' = p d p d y p d p1= 3y 2 原方程可化为d ye 2 y -1 123 1 2 12 3 1 p d p = 3y 2d y1 p 2= 2 y 2 + c2以 x = 0, y ' = p = 2, y = 1代入得c 1 = 03故由于 y ' = 3 积分得y ' = p = ±2 y 43> 0 . 故 y ' = 2 y 4，即d y= 2d xy414 y 4 = 2x + c 2以 x =0,y =1 代入得c 2 = 4y = ⎛ 1⎫4 x +1⎪故所求特解为⎝ 2 ⎭ .15. 求下列微分方程的通解：(1) y ' + y ' - 2 y = 0 ; 解：特征方程为r 2 + r - 2 = 0 解得r 1 = 1, r 2 = -2故原方程通解为(2) y ' + y = 0 ;解：特征方程为解得故原方程通解为y = c e x + c e-2x.r 2 +1 = 0 r 1,2 = ±iy = c 1 cos x + c 2 sin x(3)4 d 2 x d t 2- 20 d xd t + 25x = 0 ;解：特征方程为 4r 2 - 20r + 25 = 0r = r = 5解得1 225 t 故原方程通解为 x = (c 1 + c 2t )e 2 .(4) y ' - 4 y ' + 5 y = 0 ; 解：特征方程为解得故原方程通解为(5) y ' + 4 y ' + 4 y = 0 ;r 2 - 4r + 5 = 0 r 1,2 = 2 ± iy = e 2x (c cos x + csin x ) . 解：特征方程为 解得r 2 + 4r + 4 = 0r 1 = r 2 = -2y = e -2x (c + c x )故原方程通解为12y 11 2 x =0 x =0 ;1 2 1 2 x =0 x =0 2 ce x =0 x =0 1 2 x =0 x =0 .x ⎫ (6) y ' - 3y ' + 2 y = 0 .解：特征方程为解得故原方程通解为r 2 - 3r + 2 = 0 r = 1, r = 2y = c e x + c e 2x. 16. 求下列微分方程满足所给初始条件的特解：(1) y ' - 4 y ' + 3y = 0, y = 6, y ' = 10解：特征方程为解得 通解为r 2 - 4r + 3 = 0r 1 = 1, r 2 = 3y = c e x + c e 3xy ' = c e x + 3c e 3x⎧c 1 + c 2 = 6 ⇒ ⎧c 1 = 4 ⎨c + 3c = 10 ⎨c = 2 由初始条件得 ⎩ 1 2 ⎩ 2故方程所求特解为y = 4e x + 2e 3x .(2)4 y ' + 4 y ' + y = 0, y = 2, y ' = 0;解：特征方程为 4r 2 + 4r +1 = 0r = r = - 1解得 1 22- 1 x通解为y = (c 1 + c 2 x )e 2y ' = ⎛ c - - x c 1 - 2 ⎪ ⎝ ⎧c 1 = 2 2 2 ⎭ ⎧c = 2 ⎪ ⇒ 1⎨c - 1 c = 0 ⎨c = 1由初始条件得 ⎩⎪ 2 2 1 ⎩ 2- 1x故方程所求特解为y = (2 + x )e 2 .(3) y ' + 4 y ' + 29 y = 0, y = 0, y ' = 15;解：特征方程为解得 r 2 + 4r + 29 = 0 r 1,2 = -2 ± 5i通解为y = e -2x (c cos5x + c sin 5x ) y ' = e -2x [(5c - 2c ) cos 5x + (-5c - 2c ) sin 5x ]2112⎧c 1 = 0 ⇒ ⎧c 1 = 0⎨⎨ 由初始条件得 ⎩5c 2 - 2c 1 = 15 ⎩c 2 = 3 故方程所求特解为y = 3e -2x sin 5x .(4) y ' + 25 y = 0, y = 2, y ' = 5解：特征方程为解得r 2 + 25 = 0 r 1,2 = ±5i1 2121 2通解为y = c 1 cos 5x + c 2 sin 5x y ' = -5c 1 sin 5x + 5c 2 cos 5x⎧c 1 = 2 ⇒ ⎧c 1 = 2⎨⎨ 由初始条件得⎩5c 2 = 5 ⎩c 2 = 1故方程所求特解为17. 求下各微分方程的通解：(1)2 y ' + y ' - y = 2e x ;y = 2 cos 5x + sin 5x .解： 2r 2+ r -1 = 0∴ r = -1, r = 11 22得相应齐次方程的通解为y = c e -1x x+ c e 212令特解为 y * = A e x,代入原方程得解得 A = 1 , 故 y * = e x ,2 A e x + A e x - A e x = 2e x ,x故原方程通解为y = ex+ c e -x + c e 2.(2)2 y ' + 5 y ' = 5x 2 - 2x -1; 解： 2r 2+ 5r = 0r = 0, r = - 51 22- 5x对应齐次方程通解为y = c 1 + c 2e2令 y * = x (ax 2+ bx + c ) , 代入原方程得2(6ax + 2b ) + 5(3ax 2 + 2bx + c ) = 5x 2 - 2x -1比较等式两边系数得a = 1 ,b = - 3 ,c = 73 5 25 y * = 1 x 3 - 3 x 2 + 7 x则- 5x = 3 5 25 ⎛ 1 3 7 ⎫ y c + c e 2 + x 3 - x 2+ x ⎪故方程所求通解为12 ⎝3 5 25 ⎭ .(3) y ' + 3y ' + 2 y = 3x e -x ;解： r 2+ 3r + 2 = 0r 1 = -1, r 2 = -2 ,对应齐次方程通解为y = c e -x + c e -2x 令 y * = x ( Ax + B )e -x代入原方程得(2 Ax + B + 2 A )e -x = 3x e -x⎪ 1 2 12A = 3, B = -3解得 2 y * = ⎛ 3x 2 - 3x ⎫e -x则⎝ 2 ⎭ y = c e -x + c e -2x+ ⎛ 3 x 2 - 3x ⎫e -x1 2⎪ 故所求通解为⎝ 2 ⎭ .(4) y ' - 2 y ' + 5 y = e x sin 2x ;解： r 2- 2r + 5 = 0r 1,2 = 1± 2i相应齐次方程的通解为y = e x (c cos2x + c sin 2x ) 令 y * = x e x( A cos 2x + B sin 2x ) ，代入原方程并整理得4B cos 2x - 4 A sin 2x = sin 2xA = - 1, B = 0得4 y * = - 1x e x cos 2x则4 y = e x(c cos 2x + c 故所求通解为(5) y ' + 2 y ' + y = x ;解： r 2+ 2r +1 = 0r 1,2 = -1sin 2x ) - 1 x e xcos 2x 4 . y = (c + c x )e -x相应齐次方程通解为12令 y * = Ax + B 代入原方程得得则y = (c 2 A + Ax + B = xA = 1,B = -2y * = x - 2+ c x )e -x + x - 2故所求通解为12(6) y ' - 4 y ' + 4 y = e 2x .解： r 2- 4r + 4 = 0r 1,2 = 2y = (c + c x )e 2x对应齐次方程通解为12令 y *= Ax 2e 2 x代入原方程得2 A = 1,A = 1 2y = (c + c x )e 2 x + 1x 2e 2 x故原方程通解为1 2 2 .18. 求下列各微分方程满足已给初始条件的特解：x =π x =π ;1 2 2 (1) y ' + y + sin 2x = 0, y = 1, y ' = 1解：特征方程为得对应齐次方程通解为r 2 +1 = 0 r 1,2 = ±iy = c 1 cos x + c 2 sin x令 y *= A cos 2x + B sin 2x 代入原方程并整理得-3A cos 2x - 3B sin 2x = -sin 2xA = 0, 得B = 13 y = c 1 cos x + c 2故通解为sin x + 1 sin 2x 3 . ⎧-c 1 = 1 ⎧c 1 = -1 ⎪ ⎨-c + 2 = 1⎪ ⎨c = - 1将初始条件代入上式得⎩⎪ 2 3 ⎩⎪ 2 3 故所求特解为 y = -cos x - 1 sin x + 1sin 2x3 3 . (2) y ' -10 y ' + 9 y = e 2 x , y 解： r 2-10r + 9 = 0r 1 = 1, r 2 = 9x =0= 6 , y ' 7 x =0 = 33 7 .对应齐次方程通解为y = c e x + c e 9xA = - 1令y * = A e 2 x ,代入原方程求得 7 y = - 1e 2 x + c e x + c e 9 x则原方程通解为7 1 2c = 1 , c = 1 1 2由初始条件可求得故所求特解为y = 1 (e x + e 9 x ) - 1 e 2 x2 7 . *19. 求下列欧拉方程的通解：(1)x 2 y ' + xy ' - y = 0 解：作变换 x = e t，即 t =ln x ,原方程变为D (D -1) y + Dy - y = 0 d 2 y - =即d t 2y 0 特征方程为r 2 -1 = 0 r 1 = -1,r 2 = 1 y = c e -t + c e t = c 1+ c x 故1 2 1x 2 .(2)x 2 y ' + xy ' - 4 y = x 3 .⇒ 21 2 解：设 x = e t，则原方程化为D (D -1) y + Dy - 4 y = e 3t特征方程为r 2 - 4 = 0d 2 y -d t 2 4 y =e 3t ①故①所对应齐次方程的通解为r 1 = -2,r 2 = 2y = c e -2t + c e 2t又设 y * = A e 3t为①的特解，代入①化简得9 A - 4 A = 1 A = 1 5 ,y * = 1 e 3t5 y = c e -2t + c e 2t + 1 e 3t = c x -2 + c x 2 + 1x 3.故1 2 5 1 25。