A: $y=Ce^{4x}$
B: $y=C_1x+C_2e^{4x}$
C: $y=(C_1+C_2x)e^{4x}$
D: $y=C_1+C_2e^{4x}$
举一反三
- (4)()#include()extern()serial_initial()();()main()(){()int()a,b();()unsigned()int()x,y();()serial_initial()();()a=b=0xaa55();x=y=0xaa55;()printf("\n()a=%4x()b=%4x()x=%4x()y=%4x",a,b,x,y)();()a=a<<1();b=b>>1;()x=x<<1();y=y>>1;()printf("\n()a=%4x()b=%4x()x=%4x()y=%4x",a,b,x,y)();()printf("\n")();()printf("\n")();()printf("That()is()all.\n")();()while(1)();()}()______________________________________________________________________________________________________________________________
- 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
- 已知\( y = {x^2}{e^{ - x}} \)，则\( y'' \)为（ ）. A: \( 2{e^{ - x}} - 4x{e^{ - x}} - {x^2}{e^{ - x}} \) B: \( 2{e^{ - x}} - 4x{e^{ - x}} + {x^2}{e^{ - x}} \) C: 0 D: \( 2{e^{ - x}} - 4x{e^{ - x}} \)
- 已知\( y = {x^2} + 4x \)，则\( dy \)为（ ）. A: \( (2x + 4)dx \) B: \( 2xdx \) C: \( ({x^2} + 4)dx \) D: \( ({x^2} + 4x)dx \)
- 求下列函数的导数 (1)y＝(x²-1)³ (2)y＝cos³4x (3)y＝ln(lnx) (4)y＝arcsin(1/x)
内容
- 0
方程${{x}^{2}}{{y}^{''}}-(x+2)(x{{y}^{'}}-y)={{x}^{4}}$的通解是( ) A: $y={{C}_{1}}x+{{C}_{2}}{{e}^{x}}-(\frac{1}{2}{{x}^{3}}+{{x}^{2}})$ B: $y={{C}_{1}}x+{{C}_{2}}{{e}^{x}}-(\frac{1}{2}{{x}^{3}}+{{x}^{4}})$ C: $y={{C}_{1}}x+{{C}_{2}}x{{e}^{x}}-(\frac{1}{2}{{x}^{3}}+{{x}^{4}})$ D: $y={{C}_{1}}x+{{C}_{2}}x{{e}^{x}}-(\frac{1}{2}{{x}^{3}}+{{x}^{2}})$
- 1
问题1:准线方程为x=2的抛物线的标准方程为( ) A: y²=-4x B: y²=-8x C: y²=4x D: y²-6x=0
- 2
\( y = 2 - x \) ，\( {y^2} = 4x + 4 \) 所围图形面积为64。
- 3
求解方程组[img=218x63]1803072f0e0e849.png[/img]接近 (2,2) 的解 A: FindRoot[{x^2+y^2==5Sqrt[x^2+y^2]-4x,y==x^2},{x,2},{y,2}] B: NSolve[{x^2+y^2==5Sqrt[x^2+y^2]-4x,y==x^2},{x,2},{y,2}] C: FindRoot[{x^2+y^2==5Sqrt[x^2+y^2]-4x,y==x^2},{x,y},{2,2}] D: FindRoots[{x^2+y^2=5Sqrt[x^2+y^2]-4x,y=x^2},{x,2},{y,2}]
- 4
求解方程组[img=218x63]1803072e5daced1.png[/img]接近 (2,2) 的解 A: NSolve[{x^2+y^2==5Sqrt[x^2+y^2]-4x,y==x^2},{x,2},{y,2}] B: FindRoot[{x^2+y^2==5Sqrt[x^2+y^2]-4x,y==x^2},{x,2},{y,2}] C: FindRoot[{x^2+y^2==5Sqrt[x^2+y^2]-4x,y==x^2},{x,y},{2,2}] D: FindRoots[{x^2+y^2=5Sqrt[x^2+y^2]-4x,y=x^2},{x,2},{y,2}]