已知初值条件\( y(0) = - 4 \)，则方程\( y' + 5y = - 4{e^{ - 3x}} \)的通解中常量\( C = \)( )。______
举一反三
- 已知初值条件\( y(0) = 1 \)，则此时\( y' + y = {x^2}{e^{ - x}} \)的通解中常量\( C = \)( )。______
- 已知3个点A(x，5)，B(-2，y)，C(1，1)，若点C是线段AB的中点，则______． A: x=4，y=-3 B: x=0，y=3 C: x=0，y=-3 D: x=-4，y=-3 E: x=3，y=-4
- 求下列微分方程的通解：(1)y〞－2yˊ＝0；(2)y〞－3yˊ＋2y＝0；(3)y〞＋4y＝0；(4)y〞－4yˊ＋5y＝0；(5)y〞－6yˊ＋9y＝0；(6)y〞＋2yˊ＋ay＝0；(7)y〞＋6y〞＋10yˊ＝0；(8)y(4)－2y〞＋y＝0；(9)y(4)＋2y〞＋y＝0；(10)y(4)＋3y〞－4y＝0．
- 函数\(z = \ln \left( {3x + {y^4}} \right)\)的全微分为 A: \(dz = { { 3 + {y^4}} \over {3x + {y^4}}}dx + { { 3x + 4{y^3}} \over {3x + {y^4}}}dy\) B: \(dz = {3 \over {3x + {y^4}}}dx + { { 4{y^3}} \over {3x + {y^4}}}dy\) C: \(dz = {3 \over {3x + {y^4}}}dy + { { 4{y^3}} \over {3x + {y^4}}}dx\) D: \(dz = {3 \over {3x + {y^4}}}dx - { { 4{y^3}} \over {3x + {y^4}}}dy\)
- 如下C程序的输出是什么？#include [stdio.h]void Func1 (int x, int y);void Func2 (int *x, int *y); int main() { int x = 3; int y = 4;Func1 (x, y); printf ("x = %d, y = %d\n", x, y);Func2(&x, &y); printf ("x = %d, y = %d\n", x, y);} void Func1 (int x, int y) { x = x + y; y = x - y; x = x - y; printf ("x = %d, y = %d\n", x, y);} void Func2 (int *x, int *y) { *x = *x + *y; *y = *x - *y; *x = *x - *y;;} A: x = 3, y = 4x = 3, y = 4x = 3, y = 4 B: x = 4, y = 3x = 4, y = 3x = 4, y = 3 C: x = 3, y = 4x = 3, y = 4x = 4, y = 3 D: x = 4, y = 3x = 3, y = 4x = 4, y = 3