A: exp{itμ-σ2t2/2}
B: exp{itμ+σ2t2/2}
C: exp{2itμ-σ2t2/2}
D: exp{2itμ+σ2t2/2}
举一反三
- 求微分方程[img=269x55]17da6536a9fba07.png[/img]的通解; ( ) A: (C15*sin(2*t))/exp(3*t) + (C16*sin(2*t))/exp(3*t) B: (C15*cos(2*t))/exp(3*t) - (C16*sin(2*t))/exp(3*t) C: (C15*cos(2*t))/exp(3*t) + (C16*cos(2*t))/exp(3*t) D: (C15*cos(2*t))/exp(3*t) + (C16*sin(2*t))/exp(3*t)
- 一阶常微分方程[img=152x26]1802e4d6075ee4f.png[/img]的通解为 A: sin(2*t)/5-cos(2*t)/10+C*exp(-4*t) B: sin(2*t)/7+cos(2*t)/5-C*exp(-3*t) C: sin(2*t)/7-C*cos(2*t)/10+C*exp(-2*t) D: sin(2*t)/7-cos(2*t)/7+C*exp(-5*t)
- 用Matlab求解常微分方程初值问题[img=191x61]1802e4db6ff00c5.png[/img],输出结果是: A: 2*exp(t)+4*t*exp(-t)+1 B: 2*exp(-t)+4*t*exp(-t)-1 C: 2*exp(-t)+4*t*exp(-t)+1 D: 2*exp(t)+4*t*exp(-t)-1
- 设随机变量X服从标准正态分布N(0,1), 则E(exp(X))= A: 1 B: exp(1/2) C: exp(-1/2) D: 0
- 求微分方程[img=101x35]17da5f15503f795.png[/img] 的通解,实验命令为(). A: dsolve(Dy+2*x*y=x*exp(-x^2))ans=C1*exp(-x^2) + (x^2*exp(-x^2))/2 B: dsolve('Dy+2*x*y=x*exp(-x^2)','x')ans=C1*exp(-x^2) + (x^2*exp(-x^2))/2 C: dsolve('Dy+2*x*y=x*exp(-x^2)')ans=C1*exp(-x^2) + (x^2*exp(-x^2))/2
内容
- 0
函数[img=79x27]180355ae2690a03.png[/img]在x=2处的二阶泰勒展开式为 A: exp(sin(2))+cos(2)*exp(sin(2))*(x-2)+exp(sin(2))*(sin(2)/2-cos(2)^2/2)*(x-2)^2 B: exp(sin(2))+cos(2)*exp(sin(2))*(x-2)-exp(sin(2))*(sin(2)/2-cos(2)^2/2)*(x-2)^2 C: exp(sin(2))+cos(2)*exp(sin(2))*(x-2)-exp(sin(2))*(sin(2)/2+cos(2)^2/2)*(x-2)^2 D: exp(sin(2))+cos(2)*exp(sin(2))*(x-2)+exp(sin(2))*(sin(2)/2+cos(2)^2/2)*(x-2)^2
- 1
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))
- 2
求微分方程组[img=211x69]17da6536b48167e.png[/img]的特解Y= ( ) 。 A: 4*exp(t) B: 2*exp(t) C: exp(t) D: 8*exp(t)
- 3
求微分方程[img=372x60]17da65376dc1787.jpg[/img]的通解。 ( ) A: C26*exp(3*x) + (x*exp(3*x)*(x + 1)^2)/2 + C27*x*exp(3*x) - (x^2*exp(3*x)*(2*x + 3))/6 B: C26*exp(3*x) + C27*x*exp(3*x) - (x^2*exp(3*x)*(2*x + 3))/6 C: C26*exp(3*x) + (x*exp(3*x)*(x + 1)^2)/2 D: C27*x*exp(3*x) - (x^2*exp(3*x)*(2*x + 3))/6
- 4
10、1台液压设备中的2个元件在逻辑图上是并联的,假设两元件的失效率分别为λ,则该系统的可靠度为( )。 A: 2exp(-λt)-exp(-2λt) B: 2exp(-λt) C: exp(-2λt)【参考答案】解析:单个元件可靠度为R=exp(-λt),2个并联,则系统可靠度为=1-(1-R)^2=2R-R^2 D: 2λ