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exam1sampleans

# exam1sampleans - Sample questions for Exam 1-Answers 1...

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Sample questions for Exam 1—Answers 1. Verify that 1 x + 1 is a solution to the equation (1 + x ) y + (1 - x ) y - y = 0. Solution: y = 1 x + 1 y = - 1 ( x + 1) 2 y = 2 ( x + 1) 3 Substituting, (1 + x ) y + (1 - x ) y - y = (1 + x ) 2 (1 + x ) 3 + (1 - x ) - 1 (1 + x ) 2 - 1 1 + x = 2 (1 + x ) 2 + x - 1 (1 + x ) 2 - 1 1 + x = 2 + ( x - 1) - (1 + x ) (1 + x ) 2 = 0 . 2. Find the general solution to dy dx = xe x + y . Solution: Write xe x + y = xe x e y . Then separating variables, e - y dy = xe x dx - e - y = xe x - e x + C e - y = e x - xe x - C y = - ln ( e x - xe x - C ) (Since C is an arbitrary constant, - ln( e x - xe x + C ) would also be correct.) 3. Solve the initial value problem dy dt = y 2 - y 2 cos t , y (0) = 3. (Note: there was a typo on the questions sheet: cos x should be cos t .) Solution: Separating variables, dy y 2 = (1 - cos t ) dt - 1 y = t - sin t + C y = - 1 t - sin t + C y (0) = 3 = C = - 1 3 y ( t ) = - 1 t - sin t - 1 3 or y ( t ) = 3 3 sin t - 3 t + 1 1

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4. Find the general solution to dy dt = y t + 1. Solution: This is a first-order linear equation. In standard form, dy dt - 1 t y = 1 . Look for an integrating factor μ ( t ) such that μ = - 1 t μ : μ = - dt t ln μ = - ln t μ = e - ln t = 1 t . (We don’t need absolute value signs or constants of integration because we only need one μ that works) 1 t y - 1 t 2 y = 1 t 1 t y = 1 t 1 t y = ln | t | + C y ( t ) = t ln | t | + Ct 5. Find the solution to the initial value problem dy dx = - 2 y + xe - x , y (0) = 2. Solution: This is a first-order linear equation. In standard form, dy dx + 2 y = xe - x . We need an integrating factor μ with μ = 2 μ ; thus μ = e 2 x . e 2 x dy dx + 2 e 2 x y = xe x ( e 2 x y ) = xe x e 2 x y = xe x - e x + C y ( x ) = xe - x - e - x + Ce - 2 x Substituting x = 0 and y = 2, 2 = - 1 + C C = 3 y ( x ) = xe - x - e - x + 3 e - 2 x 2
6. Find the general solution to y - 4 y - 5 y = 0. Solution: Characteristic equation: r 2 - 4 r - 5 = 0. Characteristic roots: r = - 1 , r = 5. The general solution is y ( t ) = C 1 e - t + C 2 e 5 t . 7. Find the general solution to y + 2 y - 8 y = 0. Solution: Characteristic equation: r 3 + 2 r 2 - 8 r = 0. Characteristic roots: r = 0, r = 2 , r = - 4. The general solution is y ( t ) = C 1 + C 2 e 2 t + C 3 e - 4 t . 8. Solve the initial value problem y - 6 y + 8 y = 0 , y (0) = 3 , y (0) = - 2 . Solution: Characteristic equation: r 2 - 6 r + 8 = 0 Characteristic roots: r = 2, r = 4. The general solution is y ( t ) = C 1 e 2 t + C 2 e 4 t . Substituting in the initial values, y = C 1 e 2 t + C 2 e 4 t C 1 + C 2 = 3 y = 2 C 1 e 2 t + 4 C 2 e 4 t 2 C 1 + 4 C 2 = - 2 Solving, C 1 = 7 and C 2 = - 4.

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exam1sampleans - Sample questions for Exam 1-Answers 1...

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