35_pdfsam_math 54 differential equation solutions odd

35_pdfsam_math 54 differential equation solutions odd - x 2...

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Exercises 2.2 25. By separating variables we obtain (1 + y ) 1 dy = x 2 dx . Integrating yields ln | 1+ y | = x 3 3 + C. (2.2) Substituting y =3and x = 0 from the initial condition, we get ln 4 = 0 + C , which implies that C = ln 4. By substituting this value for C into equation (2.2) above, we have ln | 1+ y | = x 3 3 +ln4 . Hence, e ln | 1+ y | = e ( x 3 / 3)+ln 4 = e x 3 / 3 e ln 4 =4 e x 3 / 3 1+ y =4 e x 3 / 3 y =4 e x 3 / 3 1 . We can drop the absolute signs above because we are assuming from the initial condition that y is close to 3 and therefore 1 + y is positive. 27. (a) The diFerential equation dy/dx = e x 2 separates if we multiply by dx . We integrate the separated equation from x =0to x = x 1 to obtain x 1 Z 0 e
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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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