First-Order Differential Equations 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

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

2.2 Separable Variables Introduction: Separable Equations A first-order DE of the form dy/dx = g(x)h(y) is said to be separable. DEFINITION 2.1 Separable Equations

Rewrite the above equation as (2) where p(y) = 1/h(y).

(4) Integrating both sides, we have

Example 2 Solve Solution: 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, because y(4)=-3.

Fig2.18

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.

2.3 Linear Equations Introduction: Linear DEs are friendly to be solved. We can find some smooth methods to deal with. A first-order DE of the form a1(x)(dy/dx) + a0(x)y = g(x) (1) is said to be a linear equation in y. DEFINITION 2.2 Linear Equations

Standard Form Standard form of a first-order DE can be written as Standard Form Standard form of a first-order DE can be written as dy/dx + P(x)y = f(x) (2)

Solving Procedures If (2) is multiplied by (5) then (6) or (7) Integrating both sides, we get (8) Dividing (8) by gives the solution.

Integrating Factor We call as an integrating factor and we should only memorize this to solve problems.

Example 1 Solve dy/dx – 3y = 6. Solution: Since P(x) = – 3, we have the integrating factor is then is the same as So e-3xy = -2e-3x + c, a solution is y = -2 + ce3x, - < x < .

Notes The DE of example 1 can be written as y = –2 is included in the general solution. The general solution of linear first order DE include all the solutions.

Application to Circuits See Fig 2.39. (8)

Fig 2.39

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: From (8), Then Using i(0) = 0, c = -6/5, then i(t) = (6/5) – (6/5)e-20t.

Example 6 (2) A general solution of (8) is (11) When E(t) = E0 is a constant, (11) becomes (12) where the first term is called a steady-state part, and the second term is a transient term.

2.4 Exact Equations Introduction:

Differential of a Function of Two Variables If z = f(x, y), its differential or total differential is (1) Now if z = f(x, y) = c, (2) eg: if x2 – 5xy + y3 = c, then (2) gives (2x – 5y) dx + (-5x + 3y2) dy = 0 (3) Q: What is the implicit solution of (3)?

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. DEFINITION 2.3 Exact Equation

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 (4) THEOREM 2.1 Criterion for an Extra Differential

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 M(x, y) = , N(x, y) = and (why?) The sufficient part consists of showing that there is a function f for which = M(x, y) and = N(x, y)

Method of Solution Since f/x = M(x, y), we have (5) Differentiating (5) with respect to y and assume f/y = N(x, y) Then and (6) Which holds if (4) is satisfied.

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.

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

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.

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,

Example 2 (2) Thus h’(x) = 0, h(x) = c. The solution is xe2y – sin xy + y2 + c = 0

Example 3 Solve 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

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?

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.

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)

Suppose  is a function of one variable, say x, then Suppose  is a function of one variable, say x, then x = d /dx (9) becomes (10) 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 (11) In this case, if (Nx – My) / M is a function of y only, then we can solve (11) for .

We summarize the results for. M(x, y) dx + N(x, y) dy = 0 We summarize the results for M(x, y) dx + N(x, y) dy = 0 (12) If (My – Nx) / N depends only on x, then (13) If (Nx – My) / M depends only on y, then (14)

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 depends on both x and y. depends only on y. The integrating factor is e  3dy/y = e3lny = y3 = (y)

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

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 Since dy/dx = f(x, y), y = g(x, u), 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)).

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.

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

Example 2 (2) Integrating 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).

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.

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, or

Example 3 (2) then we have Solving the equation for u and the solution is or (6) Applying y(0) = 0 gives c = -1.

Example 3 (3) The graph of the particular solution is shown in Fig 2.30 in solid color.

Fig 2.30

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