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Review-Exam2Solutions

# Review-Exam2Solutions - Math136 Solutions for the Review...

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Unformatted text preview: Math136 Solutions for the Review for the 2nd exam Fall 2010 1. Find the solution of the initial value problem dy dx + 2 y x = ln x x , y (1) = 3 First find exp( R 2 /xdx ) = exp(2ln x ) = x 2 . Then multiply the equation by this: x 2 dy dx + 2 xy = x ln x The LHS is the derivative of x 2 y . We take antiderivatives on both sides to get x 2 y = R x ln xdx . Next we find the integral on the RHS by parts: u = ln x , dv = xdx , du = 1 /x , v = x 2 / 2. We obtain x 2 y = x 2 ln x 2- x 2 4 + C To find the value of C we substitute x = 1 and y = 3 into this equation. We get C = 13 / 4. So the solution is y = ln x 2- 1 4 + 13 4 x 2 2. Determine whether the following improper integrals converge or diverge, and find the values of the ones that converge. (a) Z ∞ 1 2 x √ x 2 + 1 dx (b) Z ∞ x 2 e- x dx (c) Z π/ 2 tan xdx (a) Substitute u = x 2 + 1 to find the antiderivative 2 √ x 2 + 1. Then Z ∞ 1 2 x √ x 2 + 1 dx = lim B →∞ Z B 1 2 x √ x 2 + 1 dx lim B →∞ 2 p x 2 + 1 | B 1 = lim B →∞ 2 p B 2 + 1- 2 √ 2 = ∞ This integral diverges. (b) By integrating by parts a couple of times we can get the antiderivative R x 2 e- x dx =- e- x ( x 2 + 2 x + 2) + C . Then Z ∞ x 2 e- x dx = lim B →∞ Z B x 2 e- x dx lim B →∞- e- x ( x 2 + 2 x + 2) | B = lim B →∞- e- B ( B 2 + 2 B + 2)- (- 2) We may use L’Hopital’s rule to find this last limit: lim B →∞ e- B ( B 2 + 2 B + 2) = lim B →∞ B 2 + 2 B + 2 e B L H = lim B →∞ 2 B + 2 e B = lim B →∞ 2 e B = 0 So the integral converges to 2. (c) The antiderivative is- ln(cos x ). So Z π/ 2 tan xdx = lim b → π/ 2 + Z b tan xdx lim b → π/ 2 + Z b tan xdx = lim b → π/ 2 + (- lncos( B )- (- lncos(0)) = ∞ So this one diverges. 3. Find the limits of the following sequences (if they exist): (a) lim p n 2 + 2 n- n = lim ( √ n 2 + 2 n- n )( √ n 2 + 2 n + n ) √ n 2 + 2 n + n = lim n 2 + 2 n...
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Review-Exam2Solutions - Math136 Solutions for the Review...

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