Math 293 solution set 7

# Math 293 solution set 7 - Math 293 Solutions to Problem Set...

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Math 293 Solutions to Problem Set 7. § 2.3 #3 . The equation x dx dt + t 2 x = sin t is nonlinear because of the product x dx dt . Rewrit- ing it as dx dt = sin t x - t 2 we see that it is not separable because the right-hand side cannot be factored as a product f ( x ) g ( t ). #4 . The equation is linear since it can be written e t dy dt + (ln t ) y = 3 t . It is not separable since in the form dy dt = 3 t - y ln t e t the right side cannot be factored as f ( y ) g ( t ). #15 . Rewrite the equation as y 0 + x x 2 +1 y = x x 2 +1 . To compute an integrating factor we compute R x x 2 +1 dx = ln x 2 + 1, so an integrating factor is x 2 + 1. The differen- tial equation then takes the form ( y x 2 + 1) 0 = x x 2 +1 . Integrating both sides, we get y x 2 + 1 = R x x 2 +1 dx = x 2 + 1 + C . Solving this for y gives y = 1 + C x 2 +1 . #20 . Write the equation as y 0 + 3 x y = 3 x - 2. An integrating factor is e R 3 x - 1 dx = e 3 ln x = x 3 . Multiplying both sides of the equation by x 3 then gives ( x 3 y ) 0 = 3 x 4 - 2 x 3 . Integrate both sides to get x 3 y = 3 x 5 5 - x 4 2 + C . The initial condition y (1) = 1 yields the value C = 9 10 . Solving for y , we get y = 3 x 5 5 - x 2 + 9 10 x - 3 .

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