• 2021-04-14
    Uxy= x2 y满足u(x,0)= x2,u(0,y)=cosy-1的解为1/6y2 x3+ x2+cosy-1
  • 内容

    • 0

      如下程序的运行结果是( ) intx=1,y=1;if(x==1) y=x+1;elseif(y==2) x=y+1;else y=0; A: x=1, y=2 B: x=3, y=2 C: x=3, y=0 D: x=1, y=0

    • 1

      分段函数:[img=203x91]17de5f2f296a7ac.jpg[/img],下面程序段中正确表达的是 。 A: If x >; =2 Then y = 3ElseIf x >; =1 Then y = 2ElseIf x >; =0 Then y = 1Else y = 0End If B: If x <; 0 Then y = 0If x <; 1 Then y = 1If x <; 2 Then y = 2If x >;= 2 Then y = 3 C: If x >;= 2 Then y = 3If x >;= 1 Then y = 2If x >; 0 Then y = 1If x <; 0 Then y = 0 D: If x <; 0 Then y = 0ElseIf x >; 0 Then y = 1ElseIf x >; 1 Then y = 2Else y = 3End If E: If x <; 0 Then y = 0If 0 <;= x <;1 Then y = 1If 1 <;= x <; 2 Then y = 2If x >;= 2 Then y = 3

    • 2

      ‍求解偏微分方程[img=178x28]18030731a73d552.png[/img], 应用的语句是‏ A: DSolve[(x^2+y^2)D[u,x]+x yD[u,y]==0,u,{x,y}] B: DSolve[(x^2+y^2)Dt[u[x,y],x]+xyDt[u[x,y],y]==0,u[x,y],{x,y}] C: DSolve[(x^2+y^2)D[u[x,y],x]+xyD[u[x,y],y]==0,u[x,y]] D: DSolve[(x^2+y^2)D[u[x,y],x]+xyD[u[x,y],y]==0,u[x,y],{x,y}]

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      下列函数中,( )不是方程\( xy' + y - x^2 = 0 \)的解。 A: \( y = { { {x^2}} \over 3} + {1 \over x} \) B: \( y = { { {x^2}} \over 3} \) C: \( y = { { {x^2}} \over 3} + 2 \) D: \( y = { { {x^2}} \over 3} - {1 \over x} \)

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      4.已知二元函数$z(x,y)$满足方程$\frac{{{\partial }^{2}}z}{\partial x\partial y}=x+y$,并且$z(x,0)=x,z(0,y)={{y}^{2}}$,则$z(x,y)=$( ) A: $\frac{1}{2}({{x}^{2}}y-x{{y}^{2}})+{{y}^{2}}+x$ B: $\frac{1}{2}({{x}^{2}}{{y}^{2}}+xy)+{{y}^{2}}+x$ C: ${{x}^{2}}{{y}^{2}}+{{y}^{2}}+x$ D: $\frac{1}{2}({{x}^{2}}y+x{{y}^{2}})+{{y}^{2}}+x$