First Order Differential Equations Initial Value Problems

differential equations shahbaz ahmed June 2024 First Order Differential Equations Initial Value Problems Following formulas have been used in the problems solved. log mn = n log m, m>0 ln e = 1 Proof Let ln e = z =⇒ loge e = z where e is an irrational number such that 2 < e < 3. In exponential form. ez = e Or ez = e1 =⇒ z=1 Hence ln e = 1 eln x = x Proof Let eln x = z 1 ln eln x = ln z ln x ln e = ln z Since ln e = 1 ln x = ln z Or x = z Or eln x = x Q1 dy (x + 1) dx + 2y = x, x > −1, y(0) = 1 Solution dy (x + 1) dx + 2y = x, dy dx + 2y x+1 R I.F = e x > −1, x x+1 .................... R d(x+1) 2 2 x+1 dx x+1 dx = =e eq (1) = e2 ln (x+1) 2 I.F = eln (x+1) I.F = (x + 1)2 Multiplying the equation (1) by I.F dy 2y (x + 1)2 dx + (x + 1)2 x+1 = x x+1 (x + 1)2 dy (x + 1)2 dx + 2(x + 1)y = x(x + 1) (x + 1)2 dy + 2(x + 1)ydx = x2 dx + xdx (x + 1)2 dy + yd[(x + 1)2 ] = x2 dx + xdx d[(x + 1)2 y] = x2 dx + xdx Integrating R R R d[(x + 1)2 y] = x2 dx + xdx (x + 1)2 y = x3 3 + x2 2 +C Putting x = 0 , y = 1 1= 0 3 + 0 2 +C C=1 Putting C = 1 2 y(0) = 1 (x + 1)2 y = y= 3 x3 3 + x2 2 +1 + x2 2 + 1 and y = 2 2x +3x +6 6(x+1)2 Verification Since (x + 1)2 y = x3 3 (x + 1)2 y = 3 x 3 + 2 x 2 +1 + d( x2 ) dx 2x3 +3x2 +6 6(x+1)2 are same equations.Hence Taking derivative d[(x+1)2 y] dx 3 = d( x3 ) dx 2 2 dy y d[(x+1) + (x + 1)2 dx = dx + d(1) dx 3 1 d(x ) 3 dx + 2 1 d(x ) 2 dx +0 dy 2y(x + 1) + (x + 1)2 dx = 13 (3x2 ) + 12 (2x) dy 2y(x + 1) + (x + 1)2 dx = x2 + x dy 2y(x + 1) + (x + 1)2 dx = x(x + 1) Since x > −1 =⇒ x + 1 ̸= 0 dy =x 2y + (x + 1) dx dy (x + 1) dx + 2y = x Q2 dy x dx + 2y = x2 + 1, x > 0, y(1) = 1 Solution dy x dx + 2y = x2 + 1, 2 x > 0, y(1) = 1 dy dx + x2 y = dy dx + x2 y = x + x1 ....................... eq(2) I.F=e R I.F=e2 2 x dx R dx x x x + = e2 1 x R dx x = e2 ln x = eln x 2 I.F=x2 Multiplying eq 2 by x2 3 dy x2 dx + x2 ( x2 )y = x3 + ( x1 )x2 dy + 2xy = x3 + x x2 dx x2 dy + yd(x2 ) = x3 dx + xdx d(x2 y) = x3 dx + xdx Integrating R R R d(x2 y) = x3 dx + xdx R R x2 y = x3 dx + xdx x4 4 x2 y = + x2 2 +C Putting x = 1, y = 1 1= C= 1 4 + 1 2 +C 1 4 Putting in the above equation x2 y = x4 4 + x2 4 + 1 2 y= x2 2 + 1 4 + 14 x−2 Verification Since x2 4 + 1 2 x2 y = x4 4 + y= + 14 x−2 Or x2 2 + 1 4 Taking derivative with respect to x d(x2 y) dx = 4 1 d(x ) 4 dx + 2 1 d(x )) 2 dx +0 dy x2 dx + 2xy = x3 + x Since x > 0 dy x dx + 2y = x2 + 1 4