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MATH 2270 (28)
Lecture 4

# Week 4.pdf

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School
Department
Mathematics
Course
MATH 2270
Professor
Matthew Demers
Semester
Fall

Description
1 2.6 Solving First-Order ODEs using Clever Substitutions There are several other special equations that may be solved by transforming to another variable to solve in. (Think: Integration by Substitution, but with DEs.) We cover some of them now. For all of the following examples, we will use y as the dependent variable and x as the independent variable. Substituting y(x) = v(x)x Some equations can be solved by making a substitution where we let the dependent variable be equal to some function of the independent variable multiplied by the independent variable. This can be helpful in transforming an ODE into a separable equation that can easily be solved. Example: Solve 3 2 dy = x + 2x + y : (2.106) dx x This is a non-separable ▯rst-order ODE, and in fact it can be solved by ▯nding an integrating factor. (As we saw on Assignment 1!) Let’s solve it another way, by ▯rst letting y(x) = v(x)x. Then, using the product rule, = x dv +v. Now, substituting dx dx into the equation, we obtain dv x + 2x + vx x + v = (2.107) dx x Dividing out an x from the right-hand side and bringing over the v, we get x dv = x + 2x + v ▯ v (2.108) dx In this case, the v’s cancel, and we can separate variables: dv = (x + 2)dx (2.109) 2 Integrate both sides: x 2 v = + 2x + C: (2.110) 2 y Now, recall that in our substitution, v . Back-substituting, x y x2 = + 2x + C (2.111) x 2 x3 or y = + 2x + Cx; (2.112) 2 which is the solution to the ODE. This technique is often useful for tackling ODEs with rational expressions on one side, with every term of the numerator and denominator containing the independent or dependent variables (or a mix), so that when we change \y" into \vx" we get things to cancel, simplifying the equation. If that cancellation is not going to happen, this is not the technique to use, as this substitution will just make the equation more complicated! Substituting v(x;y) = ax + by(x) + c If we have an equation of the form dy dx = f(ax + by + c); (2.113) we can transform it by substituting v(x;y) = ax + by(x) + c. This turns the DE into one that we can more easily ▯nd a solution for. Example: Solve the DE given by dy = 2 (2.114) dx x + y ▯ 3 y(e) = 1: (2.115) Let v(x;y) = x+y ▯3. Then, keeping in mind that y = y(x), dv = 1+ dy. Now, dx dx we substitute into the equation: 3 dv 2 dx ▯ 1 = v (2.116) We rearrange the equation and ▯nd common denominators so that we are ready to separate variables: dv = 2 + v (2.117) dx v Then, separating variables, we come to v dv = dx (2.118) 2 + v We can solve this by integrating both sides. (To integrate the left-hand side, use integration by substitution.) We obtain 2 + v ▯ 2lnj2 + vj = x + C: (2.119) Then, we back-substitute, remembering that v(x;y) = x + y ▯ 3. This gives us 2 + x + y ▯ 3 ▯ 2lnj2 + (x + y ▯ 3)j = x + C: (2.120) Notice that we can cancel the x terms and absorb the constants into C; then the solution is implicitly de▯ned by y ▯ 2lnjx + y ▯ 1j = C: (2.121) Applying the initial condition (2.115), (1) ▯ 2lnj(e) + (1) ▯ 1j = C (2.122) ) 1 ▯ 2lnjej = C (2.123) ) 1 ▯ 2 = C (2.124) ) C = ▯1 (2.125) Therefore, the solution to the IVP is given by y ▯ 2lnjx + y ▯ 1j = ▯1: (2.126) 4 Bernoulli Equations A very important type of di▯erential equation is a Bernoulli Equation. This is an equation of the form y + p(x)y = q(x)y : n (2.127) 1▯n This equation can be solved by making the substitution of v(x) = y . Example: Solve the equation given by t2y (t) + 2ty(t) = y ; t > 0: (2.128) This is in the form of (2.127) with n = 3. So, we make the substitution of v(t) = (y(t)) 1▯3 = y ▯2. Then, v = ▯2y ▯3y . Now, divide the equation by y . This puts the equation in a very convenient form for us to make the substitution: t y▯3 y + 2ty ▯2 = 1: (2.129) Making the substitution, we obtain ▯ t v + 2tv = 1; (2.130) 2 which is a ▯rst-order linear equation with v as the dependent variable, which we now solve. Rewriting in standard form, we get v ▯ 4t ▯1 v = ▯2t ▯2 : (2.131) We must ▯nd an integrating factor, given by R R ▯(t) = e ▯4t▯1dt = e ▯4 tdt= e ▯4 ln(t)+C= t▯4 ; (2.132) choosing C = 0 (we can drop the absolute values because t > 0). Multiply the equation by ▯(t): 5 ▯4 0 ▯5 ▯6 t v ▯ 4t v = ▯2t : (2.133) Then use the product rule backwards: d ▯4 ▯6 (t v) = ▯2t : (2.134) dt and integrate both sides with respect to t: ▯4 2 ▯5 t v = t + C: (2.135) 5 Isolate v and clean up: 2 + Ct5 v = : (2.136) 5t ▯2 Then remember that v = y , back-substitute, and solve for y, yielding the solu- tion to the equation. r 5t y = ▯ : (2.137) 2 + Ct5 2.7 Applications of First-Order ODEs Now that we can identify and solve many di▯erent types of ▯rst-order ODEs, we can use them to model various problems. Orthogonal Trajectories Two intersecting curves are orthogonal if their tangent lines are perpendicular at the point of intersection. This can be applied to a number of real-life scenarios: ▯ In electrostatics: Equipotential surfaces, which connect points with the same electrical potential, are orthogonal to the lines of an electric ▯eld. 6 ▯ In meteorology: Weather maps feature isobars along which the air pressure is constant. Orthogonal trajectories to these lines are curves depicting areas of high pressure to low pressure, and consequently wind direction. Example: Find the orthogonal trajectories to the family of curves given by x + y = C; (2.138) for constant C. This family of curves gives the set of all circles in the real-plane centred at the origin: Figure 2.1: A set of curves described by x + y + C. First, we take the equation of the circles, di▯erentiate implicitly with respect to x, and isolate fory: dx dy 2x + 2y = 0 (2.139) dx dy ▯x ) = (2.140) dx y 7 This equation tells us directly that the curves y that satisfy the equation are such dy ▯x that the slope dx, at any point (x;y), is given by y . Orthogonal trajectories will always have a perpendicular slope. From high school, we remember that a curve with a perpendicular slope to another curve will have a slope that is the negative reciprocal of the other. Thus, curves with an orthogonal trajectory must have that dy = y: (2.141) dx x This is a simple separable ▯rst-order DE. Separating the variables and solving, we obtain 1 dy = 1dx (2.142) y x lnjyj = lnjxj + C (2.143) y = cx (2.144) Thus, curves with an orthogonal trajectory are those de▯ned by y = cx, where c is a constant: T
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