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X t OÞO (s X ¯. X s X ã. Jun 17, 16eliminate the arbitrary constant y=c 1 e x c 2 xe x y'=c 1 e x c 2 (xe x e x) y''=c 1 e x c 2 (xe x 2e x) by comparing eq1 and eq,2 = eq,4 by comparing eq2 and eq3 = eq5 by comparing eq4 and eq5 = y'' 2y 3y'.

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üs WN w _ 5. (1 point) If the series y() CIxn s a solution of the differential equation ly' 5y ly 0, then cut n2 n0 cn, n 0, 1, 2 A general solution of the same equation can be written as y(x coy1(x) c1y2Ca), where 2n an x n0 2n1 bnx Calculate a2 a3. X OÞs Zs üÞ.

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W w 5 w®. üs X O ãs ã. Feb 22, 15There's always the taxicab identity $$ 12^31^3 = 10^39^3 $$ It is not known whether examples exist for all exponents For fourth powers there is $$ 133^4 134^4 = 158^4 59^4 $$ but even for fifth powers no example is known (nor a proof that there are no examples).

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X 6Þ_ üs_ W û. W!ÿ0 j7 _ 1 1d Ô. It looks as if you may have reached the right integral, which is ∫ 0 1 2 π x e − x 2 d x You can quickly integrate by making the substitution u = x 2 How to compute \int_0^1 {e^{x^2}} dx.

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6 6 s x . But then E(XjY =y)=åx xfXjY(xjy)=åx xfX(x)=E(X) 2 2 EE(g(X)jY)=E(g(X)) Proof Set Z = g(X) Statement (i) of Theorem 1 applies to any two rv’s Hence, applying it to Z and Y we obtain EE(ZjY)= E(Z) which is the same as EE(g(X)jY)= E(g(X)) 2 This property may seem to be more general statement than (i) in Theorem 1 The proof above. Ss X 6 üs Os .

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X ãs w Y w ¹. X_ O x x _Þ. Since W(t) and X(t) are both wide sense stationary and since RWX(t,τ) depends only on the time difference τ, we can conclude from Definition 1018 that W(t) and X(t) are jointly wide sense stationary Problem X(t) is a wide sense stationary random process.

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s üs Þs ã. May 31, 19Let X and Y be two independent \\mathcal{N}(0,1) random variables and Z=1XXY^2 W=1X I want to find Cov(Z,W) Solution Cov(Z,W)=Cov(1XXY^2,1X) Cov(Z,W)=Cov(XXY^2,X) Cov(Z,W)=Cov(X,X)Cov(XY^2,X) Cov(Z,W)=Var(X)E(X^2Y^2)E(XY^2)E(X) Cov(Z,W)=1E(X^2)E(Y^2)E(X)^2E(Y^2). XÞO 6 N x 6 6ss °.

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In elementary algebra, the binomial theorem (or binomial expansion) describes the algebraic expansion of powers of a binomialAccording to the theorem, it is possible to expand the polynomial (x y) n into a sum involving terms of the form ax b y c, where the exponents b and c are nonnegative integers with b c = n, and the coefficient a of each term is a specific positive. X E O â. W x X E } K W x10 N2 i10 N1 19 \ j u x X ҁv ŏI R _ C W F X g ̃G A ̕\ ɂ̓C C t g p Ă ܂ B.

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X 6Þ_ üs_ ¯Þ. T s X_ ¯. Solving the Differential equation This problem involves finding the values of the constants of a given function which satisfies the given differential equation.

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