Edwards-Penney Chapter 1, Sections 1.2, 1.3, 1.4, 1.5 Updated 2015 1.2-2 y'=(x-2)^2, y(2)=1 ===== Details: See Jennifer Lahti's solution, reference below. The METHOD OF QUADRATURE is not referenced in the textbook. See however Maple's documentation on methods of solutions for ordinary differential equations, where the first and primary method is the method of quadrature. The textbook references in maple are a record not only of the origins of the methods, but the naming conventions as well. QUADRATURE means INTEGRATE and that is the method, using the Fundamental Theorem of Calculus to evaluate integrals. The plan is to integrate both sides of the equation and determine a candidate solution to the differential equation. The textbook famously omits most steps in the examples, making it difficult to identify the method and the theory applied. See Example 1, section 1.2, for such an example. FUNDAMENTAL THEOREM OF CALCULUS Assume f is continuously differentiable and g is continuous. Then DEFINITE INTEGRALS (a) int(f'(x),x=a..b) = f(b)-f(a) (b) (d/dx)int(g(t),t=a..x)=g(x) INDEFINITE INTEGRALS (a) int(f') = f(x)+constant (b) (d/dx)int(g)=g(x) Reference: Fund Thm Calculus and method of quadrature examples http://www.math.utah.edu/~gustafso/FTC-Method-of-Quadrature.pdf Reference: Jennifer Lahti's solution to 1.2-2 http://www.math.utah.edu/~gustafso/BackgroundLogExp-decay-law-derivation.pdf 1.2-4 dy/dx = 1/x^2, y(1)=5 ===== Start with the method of quadrature to get int(dy/dx) = int(1/x^2), both indefinite integrals. Use the Fundamental theorem of calculus on the left and the power rule for integrals on the right. An answer check is expected. Either cite the back of the book or do a complete answer check, handwritten. See Jennifer Lahti's answer check to 1.2-2, above, for a sample of the expected details. 1.2-6 dy/dx = x sqrt(x^2+9), y(-4)=0 ===== Start with the method of quadrature to get int(dy/dx) = int(x sqrt(x^2+9)), both indefinite integrals. Use the Fundamental theorem of calculus on the left and u-substitution for integrals on the right. Choices for a u-substitution are not many; settle on u=x^2+9 and du=2x dx. Then int(dy/dx) = int(x sqrt(x^2+9)) = int((1/2) sqrt(u)) The last integral is evaluated with the power rule for integrals, because sqrt(u) = u^n with n=1/2. An answer check is expected. Either cite the back of the book or do a complete answer check, handwritten. See Jennifer Lahti's answer check to 1.2-2, above, for a sample of the expected details. 1.2-10 dy/dx = x exp(-x), y(0)=1 ====== Start with the method of quadrature to get int(dy/dx) = int(x exp(-x)), both indefinite integrals. Use the Fundamental theorem of calculus on the left and u-substitution for integrals on the right. int(dy/dx) = int(x exp(-x)), = int(-u exp(u)), using u=-x, du=-dx = -int(u exp(u)), a lookup in the integral table, on the back inside book cover = -((u-1)exp(u)+c, Table entry 46 = -((-x-x)exp(-x)+c, use u=-x = (x+1)exp(-x)+c Evaluate integration constant c using y(0)=1. Then check the answer. See also online solutions to 1.2-10, with more details. ===================================================================== 1.3-8 dy/dx = x^2 - y ===== Draw threaded curves on the graphic through the blue dots. The graphic is reproduced online. Reference: http://www.math.utah.edu/~gustafso/exercise1.3-8-EdwardsPenney.jpg Print a copy and draw the threaded curves, according to these rules. 1. Solution curves don't cross. 2. Threaded curves pass direction field tangent segments at nearly the same slope. 3. Threaded curves can exit the graphic top, bottom, left or right. A threaded curve y that passes through point (x0,y0) satisfies the initial condition y(x0)=y0. Reference: Lecture slides on direction fields. http://www.math.utah.edu/~gustafso/s2015/2250/Picard+DirectionFields.pdf 1.3-14 dy/dx = y^(1/3), y(0)=0 ====== The problem header is a long paragraph summarizing the Peano and Picard theorems. You are asked to test hypotheses for the two theorems, then report which theorems apply to the given problem. Reference: Lecture slides on Peano and Picard theorems. http://www.math.utah.edu/~gustafso/s2013/2250/Picard+DirectionFields.pdf PEANO THEOREM (not in the textbook, but in 1.3-14 header) Let f be continuous on a box B containing the initial point (x0,y0). Then there is a solution of y'=f(x,y) satisfying y(x0)=y0 passing edge-to-edge through a smaller box inside B. PICARD THEOREM (Theorem 1, section 1.3) Let f and its partial f_y be continuous on a box B containing the initial point (x0,y0). Then there is a solution of y'=f(x,y) satisfying y(x0)=y0 passing edge-to-edge through a smaller box inside B. The solution is unique inside the smaller box. Email Question ============== On problem 1.3-14 I'm kind of confused on what its asking for when it says the existence or uniqueness is guaranteed in some neighborhood of x=0. Do I make a domain with the center being 0,0? Is this the same equation you gave in class when you did the example y'=y^(1/3) and y(0)=0? ANSWER: Create a box B with center x=0, y=0 and verify that f(x,y)=y^(1/3) is continuous on the box. Then consider any box C with center x=0, y=0. Verify that the partial of f on y is discontinuous in C. These two checks verify that Peano's theorem applies but Picard-Lindelof fails to apply. Please refer to the Peano and Picard slides, instead of the textbook, which since the second edition has had some disastrous edits applied to Theorem 1, making problem 1.3-14 unintelligible. The original Theorem 1, to which the problem refers, contained both statements: Peano and Picard-Lindelof. ===================================================================== 1.4-2: y'+2xy=0 ====== There are two answers in the answer report. Show the steps for the equil sol y=0 and the non-equil sol y=???. An answer check is required. The easiest answer check is for the explicit sol. MAPLE ANSWER CHECK # y'+2xy=0 de:=diff(y(x),x)+2*x*y(x)=0; dsolve(de); # Answer y(x) = _C1*exp(-x^2) # Symbol _C1 is an arbitrary constant c, so we write the maple # answer as y=c exp(-x^2). It so happens that c=0 produces the # required equilibrium solution y=0, so the book's answer is # correct and maple is correct. 1.4-6: y'=3 sqrt(xy) ====== The book's answer is wrong. The problem splits into two problems (1) y' = 3sqrt(x)sqrt(y) for x>0, y>0, and (2) y' = 3sqrt(-x)sqrt(-y) for x <0, y<0 There is one equil sol y=0. Show the sol steps for the equil sol, and for each of (1),(2). That makes 3 panels of solution, which makes for 3 answers in the answer report. Check the answers to (1), (2). The equil sol y=0 has reversible steps (document your ans check accordingly). Ans equilibrium sol: y=0 Ans quadrant I: B.o.B. ans y = (x^(3/2)+c1)^2 Ans quadrant III: y = -((-x)^(3/2)+c2)^2 MAPLE ANSWER CHECK # y'=3 sqrt(xy) de:=diff(y(x),x)=3*sqrt(x*y(x)); dsolve(de,y(x)); # What a mess! solve(%,y(x)); # The answer was implicit, got 2 answers # but the equilibrium solution y=0 was skipped # Answers? Only 2 from maple, each nearly unintelligible. 1.4-8: y'=2x sec(y) ====== The separable form is cos(y)y'=2x Equil solutions arise from 2x sec(y)=0 which has no solution y=K=constant. The answer report has one answer, which is implicit. An answer check for the implicit solution is expected. WRONG ANSWERS. An explicit solution y=Arcsin(x^2+c) has been found by many people, but it is not equivalent to the implicit solution: the answer check only works with restrictions on x,c. To understand the difficulty, consider the equation sin(u+4Pi)=0.67 to be solved for u>0. Do you get the answer u+4Pi=Arcsin(0.67)? Does not this equation imply that u+4Pi is between -Pi/2 and Pi/2? The symbol arcsin (lower case A) has a different meaning. The mathematical symbols for inverse trig functions are problematic: the "arcsin" on a calculator is actually Arcsin. MAPLE ANSWER CHECK # y'=2x sec(y) de:=diff(y(x),x)=2*x*sec(y(x)); dsolve(de,y(x)); # Answer y(x) = arcsin(x^2+2*_C1) # Maple's answer uses arcsin, not Arcsin. The argument of arcsin # is not restrained to -Pi/2 to Pi/2. # The DE has no equilibrium solution, so maple is correct. 1.4-10: (1+x^2)y' = (1+y)^2 ======= The book's answer is wrong, it is missing a term of -1. The correct answer is two equations, y = -1 and y = -1 + (x+1)/(1+(x+1)c). An answer check is required for the second solution above. The equilibrium solution y=-1 answer check consists of the phrase "reversible steps." MAPLE ANSWER CHECK # (1+x^2)y' = (1+y)^2 de:=(1+x^2)*diff(y(x),x)=(1+y(x))^2; dsolve(de,y(x)); # Answer y(x) = -(arctan(x)+1+_C1)/(arctan(x)+_C1) # Dividing, y(x) = -1 + 1/(arctan(x)+_C1) # Maple skipped the answer y = -1, which cannot be obtained from # the reported maple answer. 1.4-18: x^2y' = 1 - x^2 + y^2 - x^2y^2 ======= Let y'=f(x,y), f(x,y)=(1-x^2+y^2-x^2y^2)/x^2. Then f(2,0)=-3/4 and F(x)=f(x,0)/f(2,0), G(y)=f(2,y) converts y'=f(x,y) into y'=F(x)G(y), the separable form. Divide by G(y) to get the separated form y'/G(y)=F(x), then apply quadrature. Avoid all this algebraic work by looking for the factorization y'=F(x)G(y) directly in the equation. If you find it, then F and G are known and you proceed with separation of variables, applying quadrature to the separated equation y'/G=F. Equil solutions y=K arise from G(K)=0. Solve this equation for K and report all equil sols. A full answer check is required, if the steps are not obviously reversible. The implicit answer involves arctan(y). Answer check the implicit non-equil solution, if you wish (see below). Is there an explicit solution? Yes, and you can find it by applying the tangent to each side of the implicit equation. It works, because the tan(arctan(u))=u and sec^2(u)=1+tan^2(u). Please report the explicit solution and do an answer check. MAPLE ANSWER CHECK # x^2y' = 1 - x^2 + y^2 - x^2y^2 de:=(x^2)*diff(y(x),x)=1-x^2+(y(x))^2-x^2*(y(x))^2; dsolve(de,y(x)); # Answer y(x) = -tan((x^2+1+x*_C1)/x) # There is no equilibrium solution, because # 1 - x^2 + y^2 - x^2y^2=0 is (1-x^2)(1+y^2)=0 which has no # constant solution y (a real number y that is a root). # Maple's answer happens to be correct. 1.4-22: y'=4x^3y-y, y(1)=-3 ======= There are two equations (DE+IC) and hence two answer checks. The equil sol is y=0 and since it cannot satisfy y(1)=-3, then only the non-equil solution needs to be derived and checked. The answer that results has a constant c in it, which is determined by y(1)=-3. Show all steps in obtaining the non-equil sol and do an answer check. Then find c and check that the proposed solution satisfies y(1)=-3. Is there an explicit solution? Yes, because ln|y|=expression in x. Use properties of log and exp to find the explicit solution. That formula should be reported with c evaluated, followed by an answer check of the DE and the IC. MAPLE ANSWER CHECK # y'=4x^3y-y, y(1)=-3 de:=diff(y(x),x)=4*x^3*y(x)-y(x); ic:=y(1)=-3; dsolve([de,ic],y(x)); # Answer y(x) = -3*exp(x*(x-1)*(x^2+x+1)) # By Picard's theorem, the solution is unique. Maple found it. ===================================================================== 1.5-6: Find the solution of xy'+5y=7x^2, y(2)=5 ====== The standard form of the equation is y'+(5/x)y = 7x Let W be the integrating factor W=exp(int(p(x),x)), simplify to W=x^5. Replace y'+(5/x)y by (Wy)'/W, clear fractions, apply quadrature. According to the book's answers, this gives y=x^2+c/x^5. Use y(2)=5 to obtain c=32. MAPLE ANSWER CHECK # xy'+5y=7x^2, y(2)=5 de:=x*diff(y(x),x)+5*y(x)=7*x^2; ic:= y(2)=5; dsolve([de,ic],y(x)); # Answer y(x) = (x^7+32)/x^5 1.5-10: Find the general solution of 2xy'-3y = 9x^3 ====== The standard form of the equation 2xy'-3y = 9x^3 is y'-(1.5/x)y = 4.5x^2 or y' + py = q with p(x)=-1.5/x, q(x)= 4.5x^2. Let W be the integrating factor W=exp(int(p(x),x)) and simplify to W=x^(-1.5) using n ln(x) = ln(x^n) with n=-1.5 and exp(ln(u))=u where u=x^(-1.5). Replace the standard form equation by (Wy)'/W = q. Cross-multiply and apply the method of quadrature. Then Wy=C+int(qW) reduces to the integration problem int(qW). Hence you must find int(F(x)) for F(x)=4.5(x^2)(x^(-1.5)). This is a power integral problem and the answer is (4.5/1.5) x^(1.5). Final answer: Wy=C+3x^(1.5) which simplifies to y = C x^(1.5) + 3 x^3. MAPLE ANSWER CHECK: de:=2*x*diff(y(x),x)-3*y(x) = 9*x^3; dsolve(de,y(x)); 1.5-18: Find the general solution of xy'=2y+x^3 cos(x) ====== The standard form of the equation is y'+(-2/x)y = x^2 cos(x). Let W be the integrating factor W=exp(int(p(x),x)), simplify to W=1/x^2. Replace y'+(-2/x)y by (Wy)'/W, clear fractions, apply quadrature. The book's answer is y=x^2(sin(x)+c). MAPLE ANSWER CHECK # xy'=2y+x^3 cos(x) de:=x*diff(y(x),x)=2*y(x)+x^3*cos(x); dsolve(de,y(x)); # Answer y(x) = x^2*sin(x)+x^2*_C1 # We write it as y = x^2 sin(x) + c x^2 1.5-20: Find the solution of y'=1+x+y+xy, y(0)=0 ====== The standard form of the equation is y'+(-x-1)y = x+1. Let W be the integrating factor W=exp(int(p(x),x)). Simplify to W=exp(-x^2/2-x). Replace y'+(-x-1)y by (Wy)'/W, clear fractions, apply quadrature. To integrate RHS=(x+1)W=(x+1)exp(-x^2-x), use substitution u=-x^2-x, then (x+1)W=exp(u)(-du) which integrates to -exp(u) or -exp(-x^2/2-x), which is -W! After quadrature, Wy=-W+c, and then y=-1+c/W. Use y(0)=0 to obtain c=1 and finally y = -1 + exp(x^2/2+x). The book's answer is wrong in Edwards-Penney 2/E but corrected in 3/E. The solution manual in 2/E also has an error. MAPLE ANSWER CHECK # y'=1+x+y+xy, y(0)=0 de:=diff(y(x),x)=1+x+y(x)+x*y(x); ic:= y(0)=0; dsolve([de,ic],y(x)); # Answer y(x) = -1+exp((1/2)*x*(x+2)) 1.5-22: Find the solution of y'=2xy+3x^2 exp(x^2), y(0)=5 ====== The standard form of the equation is y'+(-2x)y = 3 x^2 exp(x^2). Notation: exp(u) means e^u. For example, exp(-1) is 1/e. Let W be the integrating factor W=exp(int(p(x),x)), simplify to W=exp(-x^2). Replace y'+(-2x)y by (Wy)'/W, clear fractions, apply quadrature. Checkpoint: y=(x^3+c)exp(x^2). Use y(0)=5 to obtain c=5. MAPLE ANSWER CHECK # y'=2xy+3x^2 exp(x^2), y(0)=5 de:=diff(y(x),x)=2*x*y(x)+3*x^2*exp(x^2); ic:= y(0)=5; dsolve([de,ic],y(x)); # Answer y(x) = (x^3+5)*exp(x^2) 1.5-34: Use units of millions of cubic feet and days. ====== Reservoir volume 8000, initial pollution concentration of 0.25 percent, daily inflow 500 with pollution concentration 0.05 percent, and outflow 500. The fluid is well-mixed. Find the number of days required to reduce the concentration in the reservoir to 0.10 percent. The textbook gives the initial value problem x'=r_i c_i - (r_0/v)x, x(0)=x_0. The initial value is x_0 =(initial concentration)(volume)=(0.25/100)8000, the output rate is r_0=500, and the tank volume is V=8000. Please determine the value for the input concentration, constant c_i. You should obtain (r_i)(c_i)=1/4. Then solve the initial value problem: x'=1/4-x/16, x(0)=20, using units of millions of cubic feet and days. The answer is x(t)=4+16 exp(-t/16). This is an easy-to-solve first order linear DE with constant coefficients. The answer is x(t) = x_h + x_p, where x_h=homogeneous solution = constant/(integrating factor) and x_p is the equilibrium solution. Model Derivation Law: x'=input rate - output rate. Definition: concentration == amt/volume. Use of percentages 0.25% concentration means 0.25/100 concentration The book's answer t = 16 ln 4 = 22.2 days is correct. ===end===