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# HW1sol - Math 235 section 202 HW#1 Solution 1 dy y = te2t...

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Math 235 section 202 HW #1 Solution 1. (3 pts) Find the solution of the initial value problem: 1 2 dy dt - y = te 2 t , y (0) = 2 y 0 - 2 y = 2 te 2 t , μ ( t ) = e R - 2 dt = e - 2 t , e - 2 t y 0 - 2 e - 2 t y = 2 t , ( e - 2 t y ) 0 = 2 t , e - 2 t y = t 2 + c , c = 2, therefore y = ( t 2 + 2) e - 2 t . 2. (3 pts) Find the solution of the initial value problem: xdx + ye - x dy = 0 , y (0) = 1 xe x dx = - ydy , Z xe x dx = - Z ydy , - 1 2 y 2 = xe x - e x + c , c = 1 2 , y 2 = - 2 xe x + 2 e x - 1, y = ± - 2 xe x + 2 e x - 1, therefore, y = - 2 xe x + 2 e x - 1. (cannot be negative with given initial value.) 3. (3 pts) Find the solution of the initial value problem: y 2 (1 - x 2 ) 1 2 dy = arcsin x dx, y (0) = 1 y 2 dy = arcsin x 1 - x 2 dx , t = arcsin x , sin t = x , cos t dt = dx , 1 - x 2 = cos t , Z y 2 dy = Z tdt , 1 3 y 3 = 1 2 t 2 + c , c = 1 3 , therefore y = 3 r 3 2 arcsin 2 x + 1. 4. A tank with a capacity of 500 gallons originally contains 200 gallons of water with 100 lb of salt in solution. Water containing 1 lb of salt per gallon is entering at a rate of 3 gallons per minute, and the perfectly mixed solution flows out of the tank at a rate of 2 gallons per minute.

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HW1sol - Math 235 section 202 HW#1 Solution 1 dy y = te2t...

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