举一反三
- 求微分方程[img=364x55]17da65386dfd612.png[/img]的通解； （ ） A: - cos(2*x)*exp(x)*(x/4 - sin(4*x)/16) + C23*cos(2*x)*exp(x) + C24*sin(2*x)*exp(x) B: (3*sin(2*x)*exp(x))/32 - (sin(6*x)*exp(x))/32 - cos(2*x)*exp(x)*(x/4 - sin(4*x)/16) + C23*cos(2*x)*exp(x) + C24*sin(2*x)*exp(x) C: - sin(4*x)/16) + C23*cos(2*x)*exp(x) + C24*sin(2*x)*exp(x) D: (sin(6*x)*exp(x))/32 - cos(2*x)*exp(x)*(x/4 - sin(4*x)/16) + C23*cos(2*x)*exp(x) + C24*sin(2*x)*exp(x)
- 将函数\(f(x)=\sin^4 x\)展开成Fourier级数为 ____ . A: \(f(x) = \frac{3}{8}-\frac{1}{2}\cos 2x +\frac{1}{8}cos 4x\) B: \(f(x) = \frac{1}{4}-\frac{1}{2}\cos x +\frac{3}{8}cos 4x\) C: \(f(x) = \frac{1}{4}-\frac{1}{2}\sin 2x -\frac{3}{8}cos 4x\) D: \(f(x) = \frac{3}{8}-\frac{1}{2}\sin x -\frac{1}{8}cos 4x\)
- 常微分方程[img=243x26]1802e4d57c1aad8.png[/img]的解为： A: exp(-x)*sin(3^(1/2)*x)*C2+exp(-x)*cos(3^(1/2)*x)*C1-1/4*cos(2*x)，C1、C2为任意常数 B: exp(-2x)*cos(3^(1/2)*x)*C2+exp(-2x)*cos(3^(1/2)*x)*C1-1/4*sin(2*x)，C1、C2为任意常数 C: exp(-3x)*sin(3^(1/2)*x)*C2+exp(-3x)*sin(3^(1/2)*x)*C1-1/4*sin(2*x)，C1、C2为任意常数 D: exp(-4x)*sin(3^(1/2)*x)*C2-exp(-4x)*cos(3^(1/2)*x)*C1-1/4*cos(2*x)，C1、C2为任意常数
- 求微分方程[img=143x21]17da5f14490e50e.png[/img]的通解,实验命令为(). A: dsolve(D2y-2*Dy+5*y=sin(2*x),x)ans =exp(x)*sin(2*x)*C2+exp(x)*cos(2*x)*C1+1/17*sin(2*x)+4/17*cos(2*x) B: dsolve('D2y-2*Dy+5*y=sin(2*x)','x')ans =cos(2*x)*(sin(4*x)/17 - cos(4*x)/68 + 1/4) - sin(2*x)*(cos(4*x)/17 + sin(4*x)/68) + C1*cos(2*x)*exp(x) - C2*sin(2*x)*exp(x) C: dsolve(D2y-2*Dy+5*y=sin(2*x),'x','y')ans =exp(x)*sin(2*x)*C2+exp(x)*cos(2*x)*C1+1/17*sin(2*x)+4/17*cos(2*x)
- 求微分方程[img=634x60]17da653955cf9e7.png[/img]的特解。 （ ） A: sin(2*x)/3 - cos(x) - cos(x)/3 B: sin(2*x)/3 - cos(x) - sin(x)/3 C: cos(2*x)/3 - cos(x) - sin(x)/3 D: sin(2*x)/3 - sin(x) - sin(x)/3
内容
- 0
【单选题】设y=sin(cos(x)),求 结果为:(本题10.0分) A. cos(cos(x))*cos(x)+ sin(cos(x))*sin(x)^2 B. - cos(cos(x))*cos(x) - sin(cos(x))*sin(x)^2 C. - cos(cos(x))*cos(x)^2 - sin(cos(x))*sin(x)^2 D. - cos(cos(x))*cos(x) ^2- sin(cos(x))*sin(x)
- 1
$\int \sin^3 x \cos x dx = $ A: $\frac{\sin^4 x}{4} +C$ B: ${\sin^4 x} +C$ C: $\frac{\cos^4 x}{4} +C$ D: $\frac{\cos^4 x}{4} +C$
- 2
已知\( y = {x^3}\cos 2x \)，则\( y'' \)为（ ）. A: 0 B: \( 6x\cos 2x{\rm{ + }}12{x^2}\sin 2x - 4{x^3}\cos 2x \) C: \( 6x\cos 2x - 12{x^2}\sin 2x{\rm{ + }}4{x^3}\cos 2x \) D: \( 6x\cos 2x - 12{x^2}\sin 2x - 4{x^3}\cos 2x \)
- 3
$\int {{1 \over {3 + 5\cos x}}} dx = \left( {} \right)$ A: ${1 \over 4}\ln \left| {{{2\cos x + \sin x} \over {2\cos x - \sin x}}} \right| + C$ B: ${1 \over 4}\ln \left| {{{2\cos {x \over 2} + \sin {x \over 2}} \over {2\cos {x \over 2} - \sin {x \over 2}}}} \right| + C$ C: $\ln \left| {{{\cos {x \over 2} + \sin {x \over 2}} \over {\cos {x \over 2} - \sin {x \over 2}}}} \right| + C$ D: $\ln \left| {{{\cos x + \sin x} \over {\cos x - \sin x}}} \right| + C$
- 4
求不定积分[img=132x48]17da6537fc8dad6.png[/img]； （ ） A: -(4*(cos(x/2)/2 + 2*sin(x/2)))/(17*exp(2*x)) B: (4*(sin(x/2)/2 + 2*sin(x/2)))/(17*exp(2*x)) C: (4*(cos(x/2)/2 + 2*sin(x/2)))/(17*exp(2*x)) D: (4*(cos(x/2)/2 + 2*cos(x/2)))/(17*exp(2*x))