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CHAPTER 2 First-Order Differential Equations Contents 2.1 Solution By Direct Integration 2.2 Separable Variables 2.3 Linear Equations 2.4 Exact Equations 2.5 Solutions by Substitutions CH2_2 2.1 Solution By Direct Integration Consider dy/dx = f(x, y) = g(x). The DE dy/dx = g(x) (1) can be solved by direct integration. Integrating both sides: y = g(x) dx +c= G(x) + c. eg: dy/dx = 1 + e2x, then y = (1 + e2x) dx +c= x + ½ e2x + c CH2_3 2.2 Separable Variables Introduction: DEFINITION 2.1 Separable Equations A first-order DE of the form dy/dx = g(x)h(y) is said to be separable. CH2_4 Rewrite the above equation as p( y) dy dx g ( x) (2) where p(y) = 1/h(y). CH2_5 P ( ( x )) ' ( x ) g ( x ) (4) P ( ( x )) ' ( x ) dx Integrating both sides, we have P ( y ) dy g ( x )dx c or H ( y) G ( x) c CH2_6 Example 2 Solve Solution: dy dx ydy x , y (4) 3 y xdx and y 2 x 2 2 2 c1 We also can rewrite the solution as x2 + y2 = c2, where c2 = 2c1 Apply the initial condition, 16 + 9 = 25 = c2 See Fig2.18. Thus, y (25 x ) 2 1/ 2 because y(4)=-3. CH2_7 Fig2.18 CH2_8 Losing a Solution When r is a zero of h(y), then y = r is also a solution of dy/dx = g(x)h(y). However, this solution is not included in the general solution. That is a singular solution. CH2_9 2.3 Linear Equations Introduction: Linear DEs are friendly to be solved. We can find some smooth methods to deal with. DEFINITION 2.2 Linear Equations A first-order DE of the form a1(x)(dy/dx) + a0(x)y = g(x) is said to be a linear equation in y. (1) CH2_10 Standard Form Standard form of a first-order DE can be written as dy/dx + P(x)y = f(x) (2) CH2_11 Solving Procedures If (2) is multiplied by P ( x ) dx e then e P ( x ) dx dy e P ( x ) dx dx P ( x) y (5) e P ( x ) dx f ( x )dx (6) or P ( x ) dx d P ( x ) dx e y e f ( x) dx (7) Integrating both sides, we get P ( x ) dx e y P ( x ) dx e f ( x ) dx c (8) Dividing (8) by e P ( x ) dx gives the solution. CH2_12 Integrating Factor We call e P ( x ) dx as an integrating factor and we should only memorize this to solve problems. CH2_13 Example 1 Solve dy/dx – 3y = 6. Solution: Since P(x) = – 3, we have the integrating factor is then ( 3 ) dx 3 x e e 3 x dy 3 x 3 x e 3e y 6e dx is the same as d [e 3 x y ] 6e 3 x dx So e-3xy = -2e-3x + c, a solution is y = -2 + ce3x, - < x < . CH2_14 Notes The DE of example 1 can be written as dy 3( y 2 ) dx y = –2 is included in the general solution. The general solution of linear first order DE include all the solutions. CH2_15 Application to Circuits See Fig 2.39. L di Ri E (t ) (8) dt CH2_16 Fig 2.39 CH2_17 Example 6 Refer to Fig 2.39, where E(t) = 12 Volt, L = ½ Henry R = 10 Ohms. Determine i(t) where i(0) = 0. Solution: 1 di From (8), 10 i 12 , i ( 0 ) 0 Then 2 dt d [ e i ] 24 e 20 20 t dt i (t ) 6 ce 20 t 5 Using i(0) = 0, c = -6/5, then i(t) = (6/5) – (6/5)e-20t. CH2_18 Example 6 (2) A general solution of (8) is i (t ) e ( R / L )t L e ( R / L )t E ( t ) dt ce ( R / L )t (11) When E(t) = E0 is a constant, (11) becomes i (t ) Eo ce ( R / L )t (12) R where the first term is called a steady-state part, and the second term is a transient term. CH2_19 2.4 Exact Equations Introduction: CH2_20 Differential of a Function of Two Variables If z = f(x, y), its differential or total differential is dz f x dx f y dy (1) Now if z = f(x, y) = c, f f dy 0 dx x 3 y eg: if x2 – 5xy + y = c, then (2) gives (2x – 5y) dx + (-5x + 3y2) dy = 0 Q: What is the implicit solution of (3)? (2) (3) CH2_21 DEFINITION 2.3 Exact Equation M(x, y) dx + N(x, y) dy is an exact differential in a region R of the xy-plane, if it corresponds to the differential of some function f(x, y). A first-order DE of the form M(x, y) dx + N(x, y) dy = 0 is said to be an exact equation, if the left side is an exact differential. CH2_22 THEOREM 2.1 Criterion for an Extra Differential Let M(x, y) and N(x, y) be continuous and have continuous first partial derivatives in a region R defined by a < x < b, c < y < d. Then a necessary and sufficient condition that M(x, y) dx + N(x, y) dy be an exact differential is M N (4) y x CH2_23 Proof of Necessity for Theorem 2.1 If M(x, y) dx + N(x, y) dy is exact, there exists some function f such that for all x in R M(x, y) dx + N(x, y) dy = (f/x) dx + (f/y) dy Therefore f f M(x, y) = , N(x, y) = x y and 2 M f f f N (why?) y y x yx x y x The sufficient part consists of showing that there is a function f for which f = M(x, y) and f = N(x, y) x y CH2_24 Method of Solution Since f/x = M(x, y), we have f ( x, y ) M ( x, y ) dx g ( y ) (5) Differentiating (5) with respect to y and assume f/y = N(x, y) Then f y M ( x , y ) dx g ' ( y ) y and g ' ( y ) N ( x, y ) N ( x, y ) M ( x , y ) dx y (6) Which holds if (4) is satisfied. CH2_25 Integrate (6) with respect to y to get g(y), and substitute the result into (5) to obtain the implicit solution f(x, y) = c. CH2_26 Example 1 Solve 2xy dx + (x2 – 1) dy = 0. Solution: With M(x, y) = 2xy, N(x, y) = x2 – 1, we have M/y = 2x = N/x Thus it is exact. There exists a function f such that f/x = 2xy, f/y = x2 – 1 Then f(x, y) = x2y + g(y) f/y = x2 + g’(y) = x2 – 1 g’(y) = -1, g(y) = -y+c CH2_27 Example 1 (2) Hence f(x, y) = x2y – y+c, and the solution is x2y – y +c= c’, y = c”/(1 – x2) The interval of definition is any interval not containing x = 1 and x = -1. CH2_28 Example 2 Solve (e2y – y cos xy)dx+(2xe2y – x cos xy + 2y)dy = 0. Solution: This DE is exact because M/y = 2e2y + xy sin xy – cos xy = N/x Hence a function f exists, and f/y = 2xe2y – x cos xy + 2y that is, f ( x, y ) 2 x e xe f x e 2y 2y 2y dy x cos xydy 2 ydy sin xy y h ( x ) 2 y cos xy h ' ( x ) e 2y y cos xy CH2_29 Example 2 (2) Thus h’(x) = 0, h(x) = c. The solution is xe2y – sin xy + y2 + c = 0 CH2_30 Example 3 dy xy cos x sin x 2 , y (0) 2 Solve 2 dx y (1 x ) Solution: Rewrite the DE in the form (cos x sin x – xy2) dx + y(1 – x2) dy = 0 Since M/y = – 2xy = N/x (This DE is exact) Now f/y = y(1 – x2) f(x, y) = ½y2(1 – x2) + h(x) f/x = – xy2 + h’(x) = cos x sin x – xy2 CH2_31 Example 3 (2) We have h(x) = cos x sin x h(x) = -½ cos2 x+c Thus ½y2(1 – x2) – ½ cos2 x +c= c1 or y2(1 – x2) – cos2 x = c’ (7) where c’ = 2(c1 -c). Now y(0) = 2, so c’ = 3. The solution is y2(1 – x2) – cos2 x = 3 Q: What is the explicit solution? CH2_32 Fig 2.28 Fig 2.28 shows the family curves of the above example and the curve of the specialized IVP is drawn in color. CH2_33 Integrating Factors It is sometimes possible to find an integrating factor (x, y), such that (x, y)M(x, y)dx + (x, y)N(x, y)dy = 0 (8) is an exact differential. Equation (8) is exact if and only if (M)y = (N)x Then My + yM = Nx + xN, or xN – yM = (My – Nx) (9) CH2_34 Suppose is a function of one variable, say x, then x = d /dx (9) becomes M N d (10) y dx x N If we have (My – Nx) / N depends only on x, then (10) is a first-order ODE and is separable. Similarly, if is a function of y only, then Nx M y d (11) dy M In this case, if (Nx – My) / M is a function of y only, then we can solve (11) for . CH2_35 We summarize the results for M(x, y) dx + N(x, y) dy = 0 If (My – Nx) / N depends only on x, then ( x) e M y Nx dx N (12) (13) If (Nx – My) / M depends only on y, then ( y) e N x M y M dy (14) CH2_36 Example 4 The nonlinear DE: xy dx + (2x2 + 3y2 – 20) dy = 0 is not exact. With M = xy, N = 2x2 + 3y2 – 20, we find My = x, Nx = 4x. Since M y Nx N x 4x 2 x 3 y 20 2 2 3x 2 x 3 y 20 2 2 depends on both x and y. Nx M M y 3 y depends only on y. The integrating factor is e 3dy/y = e3lny = y3 = (y) CH2_37 Example 4 (2) then the resulting equation is xy4 dx + (2x2y3 + 3y5 – 20y3) dy = 0 It is left to you to verify the solution is ½ x2y4 + ½ y6 – 5y4 = c CH2_38 2.5 Solutions by Substitutions Introduction If we want to transform the first-order DE: dx/dy = f(x, y) by the substitution y = g(x, u), where u is a function of x, then dy dx g x ( x, u ) gu ( x, u ) du dx Since dy/dx = f(x, y), y = g(x, u), f ( x , g ( x , u )) g x ( x , u ) g u ( x , u ) du dx Solving for du/dx, we have the form du/dx = F(x, u). If we can get u = (x), a solution is y = g(x, (x)). CH2_39 Bernoulli’s Equation The DE: dy/dx + P(x)y = f(x)yn (4) where n is any real number, is called Bernoulli’s Equation. Note for n = 0 and n = 1, (4) is linear, otherwise, let u = y1-n to transform (4) into a linear equation. CH2_40 Example 2 Solve x dy/dx + y = x2y2. Solution: Rewrite the DE as a Bernoulli’s equation with n=2: dy/dx + (1/x)y = xy2 For n = 2, then y = u-1, and dy/dx = -u-2(du/dx) From the substitution and simplification, du/dx – (1/x)u = -x The integrating factor on (0, ) is e dx / x e ln x e ln x 1 x 1 CH2_41 Example 2 (2) Integrating x 1 du dx x 1 1 1 u x x x gives x-1u = -x + c, or u = -x2 + cx. Since u = y-1, we have y = 1/u and the general solution of the DE is y = 1/(−x2 + cx). CH2_42 Transformation to Separable DE A DE of the form dy/dx = f(Ax + By + C) (5) can always be transformed into a separable equation by means of substitution u = Ax + By + C. CH2_43 Example 3 Solve dy/dx = (-2x + y)2 – 7, y(0) = 0. Solution: Let u = -2x + y, then du/dx = -2 + dy/dx, du/dx + 2 = u2 – 7 or du/dx = u2 – 9 This is separable. Using partial fractions, du ( u 3 )( u 3 ) dx or 1 1 1 du dx 6 u 3 u 3 CH2_44 Example 3 (2) then we have 1 6 ln u3 u3 x c1 Solving the equation for u and the solution is or u 3 (1 ce 1 ce 6x ) 6x y 2x u 2x 3 (1 ce 1 ce 6x ) (6) 6x Applying y(0) = 0 gives c = -1. CH2_45 Example 3 (3) The graph of the particular solution y 2x u 2x 3 (1 e 1 e 6x ) 6x is shown in Fig 2.30 in solid color. CH2_46 Fig 2.30 CH2_47