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Additional info for Advanced Engineering Mathematics
Integrate to get xy = 3 4x4 +C . Then y (x) = 3 3 4x C +x for x > 0 . For the initial condition, we need 4+ C y(1) =5= so C = 17/4 and the solution of the initial value problem i s 3 17 y(x) = 4x + 4x 3 forx>0 . 6 in closed form (as a finite algebraic combination of elementary functions) . Thi s occurs with y' +xy=2 . whose general solution is y(x) = 2e-x2/2 f ex2/2 dx+ Ce -x2/2 . We cannot write f e x2/2 dx in elementary terms . 10. This provides some idea of the behavior of solutions, at least within the range of the diagram .
CHAPTER 1 First-Order Differential Equations In each of Problems 4 through 12, (a) show that the differ ential equation is not exact, (b) find an integrating factor , (c) find the general solution (perhaps implicitly defined) , and (d) determine any singular solutions the differentia l equation might have. 13. 1+ xy' = 0 ; y(e4 ) = 0 14. 3y+4xy' = 0 ; y(1) = 6 15. 2(y3 -2)+3xy2 y' = 0 ; y(3) = 1 4. xy'-3y=2x3 16. y(1 +x) +2xy' = 0 ; y(4) = 6 5. 1+(3x-e -2y )y' =0 6. 6x2 y + 12xy + y 2 + (6x2 + 2y)y' 7.
8) with M(x, y) = 2xy3 + 2 and N(x, y) = 3xZ y 2 + 8e4y . 8) can in turn be written M dx + N dy = (2xy 3 + 2) dx + (3xZ y 2 + 8e 4y) dy = 0 . Now let cP(x, y) = x2 y3 +2x+2e4y. Soon we will see where this came from, but for now, observe tha t a* = 2xy 3 + 2 = M and a* Y = 3xZ y2 + 8e 4y = N. 9) become s 0x a* dx+* dy=0 , or *hP (x, y)_°O. 9) CHAPTER 1 First-Order Differential Equation s The general solution of this equation is co(x, y) = C, or, in this example, x 2 y 3 + 2x + 2e4y = C . 8) . To verify this, differentiate the last equation implicitly with respect to x : 2xy 3 + 3x2 y2 y ' + 2 + 8e4j'y' = 0 , or 2xy3 + 2 + (3x2 y2 + 8e4j' )y ' = 0.