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Solution. Verified by Toppr. dy dx = ex −e−x. and y2 = e2x +e−2x +2. y2 −4 =e2x +e−2x −2. √y2 −4 =|ex −e−x|. and we know that for x > 0√y2 −4 =ex −e−x = dy dx. and for x <0√y2 −4 = −(ex −e−x) = −dy dx. Was this answer helpful? 27. Similar Questions. Q 1. y =ex +e−x prove that dy dx= √y2 −4. View Solution. Q 2.
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Jan 29, 2019 · 2 Answers. Sorted by: 4. As you do in your problem, we shall restrict ourselves to the discrete case, where X X and Y Y are random variables taking at most countably infinitely many possible values. Let us call this set of values S S. Let T T denote the set of products of two elements in S S.
Solve first-order linear differential equations: y' (t) - 2y (t) = 3 e^ (2t) x y' (x) - 4 y (x) = x^6 exp (x), y (1) = 0. See the steps for using Laplace transforms to solve an ordinary differential equation (ODE): solve y' (t) - 3y (t) = delta (t - 2), where y (0) = 0.
Now define covariance of X and Y by. Cov(X , Y ) = E[(X − E [X ])(Y − E [Y ]). Note: by definition Var(X ) = Cov(X , X ). Covariance (like variance) can also written a different way. Write μx = E [X ] and μY = E [Y ]. If laws of X and Y are known, then μX and μY are just constants. Then.
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Feb 9, 2020 · Let X and Y be discrete random variables. Prove the linearity of expectation E(X+Y) = E(X) + E(Y). An exercise problem in probability theory. The solution is given.
