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MAT 201E DIFFERENTIAL EQUATIONS WORKSHEET II Subject: Second Order Linear Equations 1. Find the solution of the following initial value problems and describe its behavior as t increases. (a) 6y − 5y + y = 0, y(0) = 4, y (0) = 0 (b) y + 3y = 0, y(0) = −2, y (0) = 3 2. Find a diﬀerential equation whose general solution is y = c1 e−t/2 + c2 e−2t . 3. Consider the initial value problem 2y + 3y − 2y = 0, y(0) = 1, y (0) = −β, where β > 0. (a) Solve the initial value problem. (b) Find the coordinates (t0 , y0 ) of the minimum point of the solution for β = 1. (c) Find the smallest value of β for which the solution has no minimum point. 4. Find the general solution of ty + y = 1, t > 0. (Hint: For a second order diﬀerential equation of the form y = f (t, y ), the substitution v = y , v = y leads to ﬁrst order equation of the form v = f (t, v)). 5. Determine the largest interval in which the following initial value problems are certain to have a unique diﬀerentiable solution. Do not attempt to ﬁnd the solution. (a) t(t − 4)y + 3ty + 4y = 2, y(3) = 0, y (3) = −1 (b) (x − 3)y + xy + (ln |x|)y = 0, y(1) = 0, y (1) = 1 6. If the Wronskian W of f and g is 3e4t , and if f (t) = e2t , ﬁnd g(t). 7. If f, g and h are diﬀerentiable functions, show that W (f g, f h) = f 2 W (g, h). 8. If y1 and y2 are linearly independent solutions of t2 y − 2y + (3 + t)y = 0 and if W (y1 , y2 )(2) = 3, ﬁnd the value of W (y1 , y2 )(4). 9. Solve y − 6y + 25y = 0. 10. Solve y − 8y + 16y = 0. 11. Use the method of reduction of order to ﬁnd a second solution of the (x − 1)y − xy + y = 0, x > 1; y1 (x) = ex . 12. Find a general solution of the given diﬀerential equation. (a) 2y + 3y + y = t2 + 3 sin t (b) y − y − 2y = cosh(2t) (Hint: cosh t = (et + e−t )/2) (c) y + y = 3 sin(2t) + t cos(2t) 13. Determine the form of a particular solution to y + 4y = x2 sin(2x). 14. Solve y + 6y + 9y = e−3x /x5 by using the method of variation of parameters. 15. Show that the given functions y1 and y2 satisfy the corresponding homogeneous equation; then ﬁnd a particular solution of the given nonhomogeneous equation. (a) (1 − t)y + ty − y = 2(t − 1)2 e−t , 0 < t < 1; y1 (t) = et , y2 (t) = t (b) x2 y − 3xy + 4y = x2 ln x, x > 0; y1 (x) = x2 , y2 (x) = x2 ln x

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second order, differential equations, linear differential equations, partial differential equations, linear equations, ordinary differential equations, linear differential equation, wave equation, differential equation, homogeneous equations, first order, general solution, theorem 1, hypothesis testing, second-order equations

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posted: | 3/10/2010 |

language: | English |

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