Unformatted text preview: x 2 dx = x = x 1 Z x =0 dy = y ± ± ± x = x 1 x =0 = y ( x 1 ) − y (0) . If we let t be the variable of integration and replace x 1 by x and y (0) by 0, then we can express the solution to the initial value problem as y ( x ) = x Z e t 2 dt. (b) The diFerential equation dy/dx = e x 2 y − 2 separates if we multiply by y 2 and dx . We integrate the separated equation from x = 0 to x = x 1 to obtain x 1 Z e x 2 dx = x 1 Z y 2 dy = 1 3 y 3 ± ± ± x = x 1 x =0 = 1 3 ² y ( x 1 ) 3 − y (0) 3 ³ . 31...
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 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations, Boundary value problem, initial condition, ELN, differential equation dy/dx

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