A New Equation for the Distribution of Radiant Energy by Lewis G.N.

By Lewis G.N.

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05 minutes. 8 14. When the height of the water is h, the radius of the top of the water is The differential equation is dh Ah = −c dt Aw 2 5 (20 − h) and Aw = 4π(20 − h)2 /25. 6 64h = − . 4π(20 − h)2 /25 6 (20 − h)2 Separating variables and integrating we have √ (20 − h)2 80 2 5 5 √ dh = − dt and 800 h − h3/2 + h5/2 = − t + c. 6 3 5 6 h √ Using h(0) = 20 we find c = 2560 5/3, so an implicit solution of the initial-value problem is √ √ 2560 5 80 3/2 2 5/2 5 800 h − h + h . =− t+ 3 5 6 3 √ To find the time it takes the tank to empty we set h = 0 and solve for t.

The general solution is 4ty + t2 − 5t + 3y 2 − y = c. If y(−1) = 2 then c = 8 and the solution of the initial-value problem is 4ty + t2 − 5t + 3y 2 − y = 8. 24. Let M = t/2y 4 and N = 3y 2 − t2 /y 5 so that My = −2t/y 5 = Nt . From ft = t/2y 4 we obtain f = t2 + h(y), 4y 4 3 3 t2 3 , and h(y) = − . The general solution is − 2 = c. If y(1) = 1 then c = −5/4 and the 3 2 4 y 2y 4y 2y t2 3 5 solution of the initial-value problem is − 2 =− . 4y 4 2y 4 h (y) = 25. Let M = y 2 cos x − 3x2 y − 2x and N = 2y sin x − x3 + ln y so that My = 2y cos x − 3x2 = Nx .

If y(0) = 0 then c1 = 0 and for continuity we must have c2 = 1 so that 1 1    2 − 2 (1 + x2 ) , 0 ≤ x ≤ 1; y=  3 1   − , x > 1. 3 35. We need P (x)dx = 2x, 0≤x≤1 . −2 ln x, x > 1 20 15 An integrating factor is P (x)dx e and 0≤x≤1 e2x , = 10 2 1/x , x > 1 ye2x , 0 ≤ x ≤ 1 = y/x2 , x > 1 d dx y 5 4xe2x , 0 ≤ x ≤ 1 . 4/x, x>1 3 Integrating we get 0≤x≤1 ye2x , 2 y/x , x>1 2xe2x − e2x + c1 , 0 ≤ x ≤ 1 . 4 ln x + c2 , x>1 = Using y(0) = 3 we find c1 = 4. For continuity we must have c2 = 2 − 1 + 4e−2 = 1 + 4e−2 .

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