Math 241 Linear Algebra and Ordinary Differential Equations Lecture Slides PDF

MATH 241 LINEAR ALGEBRA AND ORDINARY DIFFERENTIAL EQUATIONS LECTURE SLIDES DIFFERENTIAL EQUATIONS 2   Basic Ideas and Terminology  Separable differential...

MATH 241 LINEAR ALGEBRA AND ORDINARY DIFFERENTIAL EQUATIONS LECTURE SLIDES DIFFERENTIAL EQUATIONS 2   Basic Ideas and Terminology  Separable differential equations Homogeneous DIFFERENTIAL EQUATION  Differential Equations  First-order Linear differential equations  Bernoulli Equations  Exact Differential Equations DIFFERENTIAL EQUATION A differential equation is any equation that involves one or more derivatives of an unknown function. For example, are differential equations. In the differential equation (1.1.1) the unknown function or dependent variable is y, and x is the independent variable; in the differential equation (1.1.2) the dependent and independent variables are S and t, respectively. Differential equations such as (1.1.1) and (1.1.2) in which the unknown function depends only on a single independent variable are called ordinary differential equations. By contrast, the differential equation (Laplace’s equation) involves partial derivatives of the unknown function u(x, y) of two independent variables x and y. Such differential equations are called partial differential equations. EXAMPLE One way in which differential equations can be characterized is by the order of the highest derivative that occurs in the differential equation. This number is called the order of the differential equation.  DEFINITION EXAMPLE: The following DEs are both linear. In each case y is the dependent variable → y and its various derivatives ocur to the degree only and no products of y and/or any of its derivatives are present The major reason why it is important to study differential equations is that these types of equations pervade all areas of science, technology, engineering, and mathematics Solution of the differential equation: A solution of the differential equation is a relation between the variables which satisfies the given differential equation 𝑑𝑦 1 For example; the solution of the differential equation = 𝑥 + 2 is 𝑦(𝑥) = 𝑥 2 + 2𝑥 + 𝑐 𝑑𝑥 2 𝑑𝑦 Because it satisfies the given equation: ⇒ 𝑑𝑥 = 𝑥 + 2 putting this into the eqaution we ge ⇒ (𝑥 + 2) = 𝑥 + 2 Initial Value Problems An initial value problem, IVP, is an ordinary differential equation, ODE, with the initial point(s). The initial value helps you to find the constant c in the solution. STANDARD FORMS OF FIRST-ORDER DIFFERENTIAL EQUATIONS: First order equations may be expressed as: 𝑑𝑦 Derivative form: 𝑦 ′ = 𝑓(𝑥, 𝑦) ⇒ 𝑑𝑥 = 𝑓(𝑥, 𝑦) or Differential form: 𝑀(𝑥, 𝑦)𝑑𝑥 + 𝑁(𝑥, 𝑦)𝑑𝑦 = 0. An equation in one of these forms may be written in the other form. 𝑑𝑦 2𝑥 2 +𝑦 For example, 𝑑𝑥 = 2𝑥+4𝑦 is the first order differential equation and it can be written as (2𝑥 2 + 𝑦)𝑑𝑥 − (2𝑥 + 4𝑦)𝑑𝑦 = 0. SOLUTIONS OF THE FIRST-ORDER DIFFERENTIAL EQAUTIONS SEPARABLE DIFFERENTIAL EQUATIONS The differential equation of the form 𝑑𝑦 = 𝑓(𝑥). 𝑔(𝑦) 𝑑𝑥 is called Separable first-order differential equation.. In order to solve Separable differential equations, perform the following steps: Step 1: If possible, define the given equation as 𝑓(𝑥)𝑑𝑥 = 𝑔(𝑦). 𝑑𝑦 Step 2: Take the integral of two sides ‫)𝑦(𝑔 ׬ = 𝑥𝑑)𝑥(𝑓 ׬‬. 𝑑𝑦 to obtain 𝐺(𝑦) = 𝐹(𝑥) + 𝑐 where 𝑐 ∈ 𝑅.  EXAMPLE 1: 1 𝐴 𝐵 𝐴 𝑦 − 1 + 𝐵(𝑦 + 1) = + = 𝑦2 − 1 𝑦 + 1 𝑦 − 1 (𝑦 + 1)(𝑦 − 1) Solve the given differential equation 𝑑𝑦 𝑦 2 −1 1=𝐴 𝑦−1 +𝐵 𝑦+1 =. 𝑑𝑥 𝑥 1 If 𝑦 = 1 → 1 = 𝐴 0 + 𝐵 2 → 𝐵 = 2 Solution: 1 If 𝑦 = −1 → 1 = 𝐴 −2 + 𝐵 0 → 𝐴 = − 2 𝑑𝑦 𝑦 2 −1 1 1 = is a seperable D.E 1 −2 1 1 𝑑𝑥 𝑥 ‫ 𝑦 ׬‬2−1 𝑑𝑦 =‫׬‬ + 𝑦−1 𝑑𝑦= - 2 ln 𝑦 + 1 + 2 ln(𝑦 − 1) 2 1 1 𝑦+1 𝑑𝑦 = 𝑑𝑥 𝑦2 − 1 ณ 𝑥 𝑔(𝑦) 𝑓(𝑥) 1 1 1 1 − ln 𝑦 + 1 + ln(𝑦 + 1) = ln(𝑥) + ln 𝐶 , 𝐶>0 2 2 𝑑𝑦 = 𝑑𝑥 𝑦2 − 1 𝑥 1 1 1 𝑦+1 2 න 2 𝑑𝑦 = න 𝑑𝑥 ln = ln 𝐶𝑥 𝑦 −1 𝑥 𝑦−1 1 𝑦+1 2 = 𝐶𝑥 is the solution of the given D.E 𝑦−1 EXAMPLE 2: 𝑑𝑦 Solve + 𝑦. 𝑐𝑜𝑠 𝑥 = 0. 𝑑𝑥 Solution: 𝑑𝑦 = −𝑦. 𝑐𝑜𝑠 𝑥 is seperable 𝑑𝑥 1 𝑑𝑦 = − 𝑐𝑜𝑠 𝑥 𝑑𝑥 𝑦 1 න 𝑑𝑦 = − න cos 𝑥 𝑑𝑥 𝑦 ln 𝑦 = − sin 𝑥 + 𝑐 , 𝑐𝜖𝑅 is the solution of D.E EXAMPLE 3: 𝐿𝑒𝑡 𝑢 = sin 𝑦 2 1 Solve 𝑒 𝑥 𝑠𝑒𝑐 𝑦 𝑑𝑥 + 𝑥 𝑠𝑖𝑛 𝑦 𝑑𝑦 = 0. 𝑑𝑢 = cos 𝑦 𝑑𝑦 1 2 1 Solution: න sin 𝑦 cos 𝑦𝑑𝑦 = න 𝑢𝑑𝑢 = 𝑢 = 𝑠𝑖𝑛2 𝑦 2 1 2 2 𝑒 𝑥 𝑠𝑒𝑐 𝑦 𝑑𝑥 = − 𝑠𝑖𝑛 𝑦 𝑑𝑦 𝑥 2 𝑥2 1 𝑥2 1 1 𝑥2 𝑑𝑦 𝑒 𝑥 sec 𝑦 𝑤 න 𝑥 𝑒 𝑑𝑥 = න 2𝑥𝑒 𝑑𝑥 = න 𝑒 𝑑𝑤 = 𝑒 is seperable D.E 2 2 2 = 𝑑𝑥 1 𝑠𝑖𝑛𝑦 𝑥 1 𝑤 = 𝑥 2, 𝑑𝑤 = 2𝑥𝑑𝑥 sin 𝑦 2 sec 𝑦 = 𝑑𝑦 = 𝑥𝑒 𝑥 𝑑𝑥 cos 𝑦 sec 𝑦 2 න sin 𝑦 cos 𝑦 𝑑𝑦 = න 𝑥 𝑒 𝑥 𝑑𝑥 1 1 2 𝑠𝑖𝑛 2𝑦 = − 2 𝑒 𝑥 +c, 𝑐∈𝑅 is the solution 2  EXAMPLE 4: Solve the Initial-Value problem, IVP, 𝒆𝒙+𝒚 𝒅𝒚 − 𝒅𝒙 = 𝟎 𝒚(𝟎) = 𝟏 Solution: Separating the variables of the 𝐷𝐸 𝑒 𝑥+𝑦 𝑑𝑦 − 𝑑𝑥 = 0 we have successively 𝑒 𝑥+𝑦 𝑑𝑦 − 𝑑𝑥 = 0 , 𝑒 𝑥 𝑒 𝑦 𝑑𝑦 = 𝑑𝑥 , 𝑒 𝑦 𝑑𝑦 = 𝑒 −𝑥 𝑑𝑥 such that, by integration, we get න 𝑒 𝑦 𝑑𝑦 = න 𝑒 −𝑥 𝑑𝑥 namely 𝑒 𝑦 = −𝑒 −𝑥 + 𝑐, 𝑐 ∈ 𝑅 , Apply Initial conditions then find c 𝑒 1 = −𝑒 0 + 𝑐 ⇒ 𝑐 = 𝑒 + 1 𝑒 𝑦 = −𝑒 𝑥 + 𝑒 + 1 Sice generalsolution of the DE 𝑒 𝑥+𝑦 𝑑𝑦 − 𝑑𝑥 = 0 is 𝑦 = ln(𝑐 − 𝑒 −𝑥 ), 𝑐 ∈ 𝑅 𝑦 = ln(𝑒 + 1 − 𝑒 −𝑥 ), 𝑐 ∈ 𝑅 Separable Equations Solution: = - = - Separable Equations We use substitution method, u = x2 + 1 du = 2 xdx v = sin y 1 dv = cos ydy du = xdx 2 = - 1 1 1  2 u du = −  v dv 1 ln ( u ) + ln c = − ln ( v ) 2 ln ( x 2 + 1) + ln c = − ln ( sin y ) 1 2 Separable Equations ln ( x 2 + 1) + ln c = − ln ( sin y ) 1 2 ln ( x 2 + 1) + 2 ln sin y = 2 ln c ln ( x 2 + 1) + ln ( sin 2 y ) = ln c 2 ( ) ln ( x 2 + 1)( sin 2 y ) = ln c 2 (x 2 + 1)( sin y ) = c 2 2 or (x 2 + 1)( sin 2 y ) = c where 𝑐 = 𝑐 2 HOMOGENOUS DIFFERENTIAL EQUATIONS DEFINITION A function f (x, y) is said to be homogeneous of degree zero if f (tx, ty) = f(x, y) for all positive values of t for which (tx, ty) is in the domain of f. More generally, f (x, y) is said to be homogeneous of degree m if f (tx, ty) = 𝑡 𝑚 𝑓 (x, y). Equivalently, we can say that f is homogeneous of degree zero if it is invariant under a re-scaling of the variables x and y. The simplest nonconstant functions that are 𝑦 𝑥 homogeneous of degree zero are 𝑓 𝑥, 𝑦 = and 𝑓 𝑥, 𝑦 = 𝑥 𝑦 Thus f can be considered to depend on the single variable V = y/x. The following theorem establishes that this is a basic property of all functions that are homogeneous of degree zero. EXAMPLE: In the previous example, if we factor an 𝑥 2 term from the numerator and denominator, then the function f can be written in the form a function f (x, y) that is homogeneous of degree zero depends only on the combination y/x and hence can be considered as a function of a single variable, say, V, where V = y/x. DEFINITION If f (x, y) is homogeneous of degree zero, then the differential equation 𝑑𝑦 = 𝑓(𝑥, 𝑦) 𝑑𝑥 is called a homogeneous first-order differential equation. In general, if 𝑑𝑦 = 𝑓(𝑥, 𝑦) 𝑑𝑥 𝑑𝑦 𝑦 Then =𝐹( ) 𝑑𝑥 𝑥 𝑦 Let v = then 𝑦 = 𝑣𝑥 𝑥 𝑑𝑦 𝑑𝑣 =𝑣+𝑥 𝑑𝑥 𝑑𝑥  1 𝑣 1 EXAMPLE We write this 4(1+𝑣 2 ) − 1+𝑣 2 𝑑𝑣 = 𝑥 𝑑𝑥 Which can be integrated directly to obtain 1 1 arctan 𝑣 − ln 1 + 𝑣 2 = ln 𝑥 + 𝑐 4 2 𝑦 𝑦 𝑑𝑦 𝑥(4 + ) Substituting 𝑣 = 𝑥 and multiplying through 2 yields = 𝑥 (4 + 𝑣) 𝑑𝑣 𝑑𝑥 𝑥(1 − 4 𝑦) =𝑣+𝑥 1 𝑦 𝑥2 + 𝑦2 𝑥 (1 − 4𝑣) 𝑑𝑥 arctan − ln 2 = ln 𝑥 2 + 𝑐1 2 𝑥 𝑥 Then separate the variables 𝑦 (4 + 𝑣) 𝑑𝑣 Let v = 𝑥 → 𝑦 = 𝑣𝑥 −𝑣 =𝑥 Which simplifies to (1 − 4𝑣) 𝑑𝑥 1 𝑦 𝑑𝑦 𝑑𝑣 arctan − ln 𝑥 2 + 𝑦 2 = 𝑐1 2 𝑥 =𝑣+𝑥 4(1 + 𝑣 2 ) 𝑑𝑣 𝑑𝑥 𝑑𝑥 =𝑥 (1 − 4𝑣) 𝑑𝑥 𝑑𝑦 𝑥(4 + 𝑣) Seperating ther variables gives = 𝑑𝑥 𝑥(1 − 4𝑣) (1 − 4𝑣) 1 𝑑𝑣 = 𝑑𝑥 (4 + 𝑣) 𝑑𝑣 4(1 + 𝑣 2 ) 𝑥 =𝑣+𝑥 (1 − 4𝑣) 𝑑𝑥 EXAMPLE Solution: Check if the give DE is homogenous or not EXAMPLE: FIRST-ORDER LINEAR DIFFERENTIAL EQUATIONS Definition:  + yP( x ) = Q( x ) Notice the coefficient of dy/dx is 1 dy dx We multiply by an f ( x) + yf ( x) P( x ) = f ( x)Q( x ) dy dx integrating factor, f(x) The first term is one of the terms we get when we differentiate yf(x). We want the second term to be the other. d  yf (x ) = f (x ) + yf ' (x ) dy  P(x )dx =  f ' ( x) dx dx dx f ( x) yf ( x) P(x ) = yf ' ( x)  P(x )dx = ln f (x ) f (x ) = e  P ( x ) dx P(x ) = f ' ( x) f ( x) THE INTEGRATING FACTOR f (x ) = e  P ( x ) dx This is what we use to find the integrating factor EXAMPLE 1  ex f (x ) = e P ( x ) dx dy Note: The coefficient of dy/dx must be 1 x + 3y = 2 dx x dy 3 y e x + = 3 P(x) is the coefficient of y dx x x ( ) Px = 3 x 3  x dx f (x ) = e f ( x) = e 3 ln x =e ln x 3 = x3 This is our integrating factor dy x3 + 3x 2 y = e x x 3 y =  e x dx = e x + c dx ex + c d 3 dx ( ) x y = ex y= x3 EXAMPLE 2: dy f (x ) = e  P ( x ) dx + y cot x = cos ecx dx f (x ) = e  cot xdx = e ln sin x = sin x This is our integrating factor dy sin x + y cos x = 1 dx d ( y sin x ) = 1 dx y sin x =  1dx = x + c x+c y= sin x EXERCISE 1. Find the general solution to the differential equation dy 1 x −y= 2 dx x 2. Given that when x = 0, y = 1, solve the differential equation dy + y = 4 xe x dx EXAMPLE 3:  f (x ) = e P ( x ) dx 𝑑𝑦 2𝑥+1 Solve + 𝑦 = 𝑒 −2𝑥 Note: The coefficient of dy/dx is 1 𝑑𝑥 𝑥 2 𝑥 2𝑥 + 1 P(x) is the coefficient of y 𝑥𝑒 2𝑥 𝑦 = +𝑐 Integrating, we obtain the 𝑝 𝑥 = 2 𝑥 solutions 2𝑥+1 ‫𝑥 ׬‬ 𝑓 𝑥 =𝑒 = 𝑒 (2𝑥+ln 𝑥 ) 𝑥𝑒 −2𝑥 𝑐 −2𝑥 = 𝑒 2𝑥 𝑒 𝑙𝑛𝑥 = 𝑥𝑒 2𝑥 This is our integrating factor 𝑦 = + 𝑒 2 𝑥 𝑑𝑦 2𝑥 + 1 𝑥𝑒 2𝑥 +𝑒 2𝑥 𝑦 = 𝑥𝑒 2𝑥 𝑒 −2𝑥 𝑑𝑥 𝑥 𝑑𝑦 𝑥𝑒 2𝑥 + 𝑒 2𝑥 2𝑥 + 1 𝑦 = 𝑥 𝑑𝑥 𝑑 𝑥𝑒 2𝑥 𝑦 = 𝑥 𝑑𝑥 ANSWERS 1. Find the general solution to the differential equation dy 1 1 x −y= 2 y = cx − dx x 2x 2. Given that when x = 0, y = 1, solve the differential equation y = (2 x − 1)e x + 2e − x dy + y = 4 xe x dx EXAMPLE 3: Apply initial conditions 𝑦 2 = 1 BERNOULLI DIFFERENTIAL EQUATIONS In this section we are going to take a look at differential equations in the form, 𝑦 ′ + 𝑝 𝑥 𝑦 = 𝑞(𝑥)𝑦 𝑛 where p(x) and q(x)are continuous functions on the interval we’re working on and n is a real number. Differential equations in this form are called Bernoulli Equations.  First notice that if n=0 or n=1 then the equation is linear and we already know how to solve it in these cases. Therefore, in this section we’re going to be looking at solutions for values of nn other than these two.  In order to solve these we’ll first divide the differential equation by 𝑦 𝑛 to get, 𝑦 −𝑛 𝑦 ′ + 𝑝 𝑥 𝑦1−𝑛 = 𝑞(𝑥)  We are now going to use the substitution 𝑣 = 𝑦 𝑛−1 to convert this into a differential equation in terms of v. As we’ll see this will lead to a differential equation that we can solve.  We are going to have to be careful with this however when it comes to dealing with the derivative, y′y. We need to determine just what y′ is in terms of our substitution. This is easier to do than it might at first look to be. All that we need to do is differentiate both sides of our substitution with respect to x. Remember that both v and y are functions of x and so we’ll need to use the chain rule on the right side. Then implicit differentiation gives us, 𝑣 ′ = (1 − 𝑛)𝑦 −𝑛 𝑦 ′ Now, plugging this as well as our substitution into the differential equation gives, Now, plugging this as well as our substitution into the differential equation gives, 1 𝑣′ + 𝑝 𝑥 𝑣 = 𝑞 𝑥 1−𝑛 This is a linear differential equation that we can solve for v and once we have this in hand we can also get the solution to the original differential equation by plugging v back into our substitution and solving for y. Let’s take a look at an example. EXAMPLE Solve EXAMPLE 1 BERNOULLI DIFFERENTIAL EQUATIONS Definition: An equation of the form 𝑑𝑦 + 𝑃(𝑥)𝑦 = 𝑄(𝑥). 𝑦 𝑛 (1) 𝑑𝑥 is called a Bernoulli equation, where 𝑛 ≠ 0,1. If 𝑛 = 0, equation (1) is the Linear equation. If 𝑛 = 1, equation (1) is the Separable equation. Bernoulli differential equation can be converted to a linear equation by using the transformation 𝑢 = 𝑦1−𝑛.  To solve Bernoulli differential equations, the following steps can be applied: 𝑑𝑦 + 𝑃(𝑥)𝑦 = 𝑄(𝑥). 𝑦 𝑛 where 𝑛 ≠ 0,1 (1) 𝑑𝑥 Equation (1) is in the standard form Step1: the derivative of the transformation. 𝑑𝑢 𝑑𝑦 The derivatives depend on 𝑢 = 𝑦 1−𝑛 ⇒ = 1 − 𝑛 𝑦 −𝑛. 𝑑𝑥 𝑑𝑥 Step2: Multiply equation (1) by (1 − 𝑛)𝑦 −𝑛 𝑑𝑦 (1 − 𝑛)𝑦 −𝑛 𝑑𝑥 + (1 − 𝑛)𝑦 −𝑛. 𝑃(𝑥)𝑦= (1 − 𝑛)𝑦 −𝑛 𝑄(𝑥). 𝑦 𝑛 𝑑𝑢 (1−𝑛)𝑃(𝑥).𝑢 (1−𝑛).𝑄(𝑥) 𝑑𝑥 Step 3: The equation in step 2 is the linear differential equation. 𝑑𝑢 We only solve the linear equation which is + 𝑃1 (𝑥)𝑢 = (1 − 𝑛). 𝑄(𝑥). 𝑑𝑥 EXAMPLE: 𝑑𝑦 2𝑦 2 1 Solve the differential equation + = 𝑦2𝑥2 𝐼=𝑒 ‫ ׬‬−𝑥𝑑𝑥 =𝑒 −2 ln 𝑥 =𝑒 ln 𝑥 2 = 2 𝑑𝑥 𝑥 𝑥 Solution: 𝑑𝑦 2 + 𝑥. 𝑦 = 𝑥. 𝑦 2 ➔𝑛=2 (1) 1 𝑑𝑢 2 1 2 𝑑𝑥 − 𝑢 = −𝑥 𝑥 2 𝑑𝑥 𝑥 𝑥 2 𝑑𝑢 𝑑𝑦 𝑢 = 𝑦 1−2 ⇒ 𝑢 = 𝑦 −1 ⇒ = −𝑦 −2. 𝑑𝑥 𝑑𝑥 1 ′ ( 𝑢) = −1 𝑥2 Multiply (1) by −𝑦 −2 , then we obtain 1 න( 2 𝑢)′ 𝑑𝑢 = න −1𝑑𝑥 𝑑𝑦 2 𝑥 −𝑦 −2. − 𝑦. (. 𝑦) = −𝑦 −2. (𝑥 2. 𝑦 2 ) −2 𝑑𝑥 𝑥 1 𝑑𝑦 2 −1 𝑢 = −𝑥 + 𝑐 −𝑦 −2. −.. 𝑦 = −𝑥 2 𝑥2 𝑑𝑥 𝑥 𝑢 = −𝑥 3 + 𝑐𝑥 2 𝑑𝑢 2 − 𝑢 = −𝑥 2 𝑦 −1 = −𝑥 3 + 𝑐𝑥 2 𝑦 = (−𝑥 3 + 𝑐𝑥 2 )−1 𝑑𝑥 𝑥 EXAMPLE 𝑑𝑦 𝑦 𝑦2 Solve the differential equation − = − It is in the standard form 𝑑𝑥 𝑥 𝑥 Solution: 𝑑𝑦 1 1 −. 𝑦 = −. 𝑦2 ➔𝑛=2 (1) 𝑑𝑥 𝑥 𝑥 𝑑𝑢 𝑑𝑦 𝑢 = 𝑦 1−2 ⇒ 𝑢 = 𝑦 −1 ⇒ = −𝑦 −2. 𝑑𝑥 𝑑𝑥 Multiply by −𝑦 −2 , then we obtain 𝑑𝑦 1 1 −𝑦 −2. − 𝑦 −2. (−. 𝑦) = −𝑦 −2. (−. 𝑦 2 ) Simplify this equation 𝑑𝑥 𝑥 𝑥  𝑑𝑦 Example 2: Solve the I.V.P. 𝑑𝑥 − 𝑦 = 𝑒 𝑥 𝑦 2 , 𝑦 0 =1 Convert this equation to linear equation This is also the solution of the IVP This is the solution to the Bernoulli equation EXACT DIFFERENTIAL EQUATIONS EXACTNES TEST Example: Solve 3𝑥 2 + 4𝑥𝑦 𝑑𝑥 + 2𝑥 2 + 2𝑦 𝑑𝑦 = 0 𝜕𝑀 𝜕𝑁 = 𝜕𝑦 𝜕𝑥 Comparing differential equation with and  𝜕𝑓 Integrating = 𝑀 𝑥, 𝑦 with respect to 𝑥 (3) 𝜕𝑥 𝑓 𝑥, 𝑦 = න 𝑀 𝑥, 𝑦 𝑑𝑥 + 𝑘 𝑦 𝑓 𝑥, 𝑦 = න 3𝑥 2 + 4𝑥𝑦 𝑑𝑥 + 𝑘 𝑦 Find 𝑓 𝑥, 𝑦 = 𝑥 3 + 2𝑥 2 𝑦 + 𝑘 𝑦 𝑘(𝑦) (4) 𝜕𝑓 𝑑𝑘(𝑦) (5) Differentiating w.r.t ‘y’ = 2𝑥 2 + 𝜕𝑦 𝑑𝑦 Example: Show that the differential equation is exact and solve the differential equation 𝑦 − 𝑥𝑦 2 + 2𝑦𝑒 𝑥 𝑑𝑥 + 𝑥 − 𝑥 2 𝑦 + 2𝑒 𝑥 𝑑𝑦 = 0 𝑥 2𝑦2 𝑓 𝑥, 𝑦 = 𝑥𝑦 − + 2𝑦𝑒 𝑥 + ℎ 𝑦 → 3 2 F ( x, y ) = c F ( 0, 2 ) = c = −2 Exercises 1. 2.