Math Differential Equations Material Covering Lab 2

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1 Math Differential Equations Material Covering Lab 2 Separable Equations A separable equation is a first-orer ODE that can be written in the form y'= g x $h y This is the general metho to solve such ifferential equation: y x = g x $h y 1 y = g x x C c (abuse of notation) h y Remark: it is enough to a the constant only on one sie of the equation. Example 1: Solve the following DE. y'c 2$x$y 2 = 0 Solution: to be one in class. Let us ouble check our answer by using Maple: oe y'c 2$x$y 2 = 0 x y x C 2 x y x 2 = 0 solve oe y x = 1 x 2 C Example 2 (Exercise 2 on Lab 2): Solve the following IVP y' = y2 K 1 x 2 y 2 = 2 K 1 Solution: to be one in class. Let us ouble check our answer by using Maple: oe y' = y2 K 1 x 2 K 1 ics y 2 = 2 solve oe, ics x y x = y x 2 K 1 x 2 K 1 y 2 = 2 y x = x Look at Example 3 Section 2.2 of your textbook. (1.1) (1.2) (1.3) (1.4) (1.5) Linear Equations We want to solve 1st-rer linear ODE, which can be written as

2 a 1 x $y'c a 0 x $y = g x Diviing both sies by a 1 x we can rewrite the DE as y' C P x $y = Q x Remark. We are looking for a solution y = f x on an interval I where P x an Q x are both (efine an) continuous. Important. When possible solve by separation of variables since it is, usually, faster. General Solution The key iea (from Johann Bernoulli) is to multiply both sies by the same function µ x : µ x $y' x C µ x $P x $y = µ x $Q x (1) Consier the function µ x $y x an take its erivative x µ x $y x = µ x $y' C µ' x $y x Notice: if we have that µ' x = µ x $P x then equation (1) becomes x µ x $y x = µ x $Q x (2) Equation (2) is very nice because we alreay know how to solve it: µ x $y x = µ x $Q x x C C Therefore, we have reuce to solve the following ODE µ x = µ$p x We solve the above equation by separation of variables: µ µ = P x x from which ln µ x = P x x C c 1

3 an µ x = c 2 $e P x x c where c 2 = e 1 Remark. It is enough to fin just one function µ x, so we can choose c 2 = 1. Therefore we have µ x = e P x x. The ifferential equation becomes x e P x x $y x = e P x x $Q x At this point we solve the problem via simple integration. See equation (8) on page 47 of your textbook for the final formula for y x. However, the only formuala we nee to memoriza is the integrating factor µ x = e P x x. Example. Solve the DE y'= 5 K 8 y K4$x$y x C 2 2 an the IVP with initial conition y K1 = 2. Solution: to be one in class. Let us ouble check our solution on Maple: oe y'= 5 K 8 y K4$x$y x C 2 2 ics y K1 = 2 solve oe solve oe, ics x y x = 5 K 8 y x K 4 x y x x C 2 2 y x = y x = y K1 = x C 2 3 C x C 2 4 x C 2 3 C 1 3 x C 2 4 (2.1) (2.2) (2.3) (2.4)

4 Exact Equations Recall. So far, we can solve : 1st-orer separable DE (not necessary linear) 1st-orer linear DE Now we focus on 1st-orer DE that can be written in the form M x, y x C N x, y y = 0 Recall from Calc III : if z = f x, y then z = vf vf x C vx vy y when f x, y = c, a constant, we have vf vf x C y = 0. vx vy 2 Example: consier the function f x, y x C 2$ sin x C y C 2$ sin y 2 2 x, y / x C 2 sin x C y C 2 sin y 2 A one-parameter family of solutions for the DE x f x, y x C f x, y y = 0 y 2 x C 2 sin x 1 C 2 cos x x C 2 y C 2 sin y 1 C 2 cos y y = 0 is given by f x, y = c. Let us use Maple to plot a special solution: f x, y = 36$p. with plots animate, animate3, animatecurve, arrow, changecoors, complexplot, complexplot3, conformal, conformal3, contourplot, contourplot3, coorplot, coorplot3, ensityplot, isplay, ualaxisplot, fielplot, fielplot3, graplot, graplot3, implicitplot, implicitplot3, inequal, interactive, interactiveparams, intersectplot, listcontplot, listcontplot3, listensityplot, listplot, listplot3, loglogplot, logplot, matrixplot, multiple, oeplot, pareto, plotcompare, pointplot, pointplot3, polarplot, polygonplot, polygonplot3, polyhera_supporte, polyheraplot, rootlocus, semilogplot, setcolors, setoptions, setoptions3, spacecurve, sparsematrixplot, surfata, textplot, textplot3, tubeplot implicitplot f x, y = 36$p, x =K15..15, y =K15..15, grirefine = 2 ; (3.1) (3.2) (3.3)

5 10 y 5 K10 K x K5 K10 Question: Given the DE M x, y x C N x, y y = 0, how can we check whether there exists f x, y such that 1 M x, y = vf vx an 2 N x, y = vf? vy Definition: we say that the DE: M x, y x C N x, y y = 0 is exact if there exists a function f x, y that satisfies properties (1) an (2). Theorem (Theorem 2, page 57 of your textbook): Suppose that the first partial erivatives of M x, y an N x, y are continuous in a rectangle R. Then M x, y x C N x, y y = 0,

6 is exact if an only if vm vy x, y = vn vx x, y hols for every x, y in R. Example/Explanation: Consier the DE: 2$x$y 2 K3 x C 2 x 2 y C 4 y = 0. Determine whether the DE is exact. If it is exact, solve it. Solution: to be one in class. See also examples 2, 3, an 4 on your textbook. Remark: to solve an IVP plug-in x an y into the final solution f x, y = c in orer to etermine c. Then write the final equation (even in implicit form). Special Integrating Factor Question: What if the DE: M x, y x C N x, y y = 0 is not exact? Answer: We look for a suitable intagraing factor. More precisely: we want to fin a function µ x, y such that the DE: µ x, y $M x, y x C µ x, y $N x, y y = 0 is exact. First of all, assume that the continuity hypothesis works. Then we want: µ M y = µ N x Let us expan both sies: µ y M C µ M y = µ x N C µ N x now we manipulate our equation to get µ x N Kµ y M = µ M y K N x (3) Simplifying assumptions:

7 Case 1) µ is only a function in terms of x. This means that µ y = 0 an from (3): µ x = M y K N x N µ We can etermine µ x if M y K N x N µ x = e M y K N x N x only epens of x: Case 1) µ is only a function in terms of y. This means that µ x = 0 an from (3): µ y = N x KM y M µ We can etermine µ y if N x KM y M only epens of x: µ y = e N x KM y M y Remark: the solutions of the new DE are almost equivalent: we are aing µ x = 0 or µ y = 0 obtaine, in general, only for finite values of xs an ys. Example/Explanation: Consier the DE: 2 y 2 C 3 x x C 2 x y y = 0. a) Check that the DE is not exact. b) Solve the DE via integrating factor. Solution: to be one in class. Substitutions an Transformations Homogeneous Equations We say that the DE y'= f x, y

8 is homogeneous if f x, y can be written as a function of y x. In this case we apply the substitution y = v$x. At the en recover y(x). Keep in min that: y' = v C x$ v x. After substituting, we can solve by separation of variables. Example: Solve the DE y 2 C y$x x C x 2 y = 0. Solution: to be one in class. Let us check our answer by using Maple: oe y 2 C y$x C x 2 $y' = 0 y x 2 C y x x C x 2 x y x = 0 solve oe y x = 2 x K1 C 2 x 2 (5.1.1) (5.1.2) Equations of the form y'=g(ax+by) When the DE can be written as y'= G a x C b y we can apply the substitution z = a x C b y. Assume b s 0 (otherwise we can immeiately apply separation of variables). We have y'= 1 b z' K a b. After substituting, we can solve by separation of variables. At the en recover y(x). Example: Solve the DE y'= x C y K1. Solution: to be one in class. Let us check our solution on Maple oe y'= solve oe x C y K1 x y x = x C y x K 1 (5.2.1)

9 x K 2 x C y x K = 0 Remark: this argument also works with y'= G a x C b y C c. (5.2.2) Bernoulli equations A Bernoulli equation is an equation that can be written in the form y' C P x y = Q x y n with n s 1. Sunstitution: v = y 1 Kn. We get y x = 1 1 K n yn v x. After substituting solve as a 1st-orer linear DE. At the en recover y(x). Example: Solve the DE y' - y = e x $y 2. Solution: to be one in class. Let us check our solution on Maple: oe y' Ky = exp x $y 2 solve oe x y x K y x = ex y x 2 y x = 2 Ke x C 2 e Kx (5.3.1) (5.3.2) Equations with linear coefficients A DE that can be written in the form a 1 x C b 1 y C c 1 x C a 2 x C b 2 y C c 2 y = 0 it is calle DE with linear coefficients. When a 1 b 2 = a 2 b 1 then the DE above can be reuce to the form y'= G a x C b y. Otherwise, we have two cases (I) c 1 = c 2 = 0 then the DE becomes homogeneous.

10 (II) apply the substitution x = u C h an y = v C k an etermine h an k in orer to reuce to case 1. Example: Solve the DE 2 x Ky x C 4 x C y K3 x = 0. Solution: to be one in class. Let us ouble check on Maple oe y'=k solve oe y x = 1 C 1 24 K 1 2 x Ky 4 x C y K 3 x y x = K 2 x K y x 4 x C y x K 3 1 K72 2 x K 1 C x K 1 2 C 8 C x 2 x K 1 K4 C 27 2 x K / 3 K x (5.4.1) (5.4.2) K 1 K 1 K72 2 x K 1 C x K 1 2 C 8 C x K 1 2 x K 1 K4 C 27 2 x K / 3 K C 3 2 x K 1 K 1 4 $I K72 2 x K 1 C x K 1 2 C 8 C x K 1 2 x K 1 K4 C 27 2 x K / 3 C x K 1 K 1 K72 2 x K 1 C x K 1 2 C 8 C x K 1 2 x K 1 K4 C 27 2 x K / 3