MATH 315 Differential Equations (Fall 017) Homework 3 Solutions(Part ) Due Frida Sept. 8 Part : These will e graded in detail. Be sure to start each of these prolems on a new sheet of paper, summarize the prolem, and explain what ou are doing so that a classmate who is struggling can follow what ou are doing just reading our work (do not refer to an external references, including the text). 1. (10 pts) Solve the following initial value prolem using the Integrating Factor Method hand Check to e sure our answer is correct. (x + 1) + 3x = 6x, (0) = 1 We want to solve the first-order ODE aove using the Integrating Factor Method. Since x is the independent variale, we want to e sure the equation can e written in the form: + p(x) = g(x) first. Naturall, we can isolate dividing oth sides x + 1 to otain the equation: + ( 3x ) = 6x. With p(x) = 3x and g(x) = 6x, we are read to find our Integrating x +1 x +1 x +1 x +1 Factor µ(x) = e p(x)dx. p(x)dx = 3x dx. Appling a u-sustitution with u = x +1 x + 1, du = xdx and therefore p(x)dx = 3 du = 3 ln u = 3 ln u x + 1 since 1 du = ln u + C and for the integrating u factor, the aritrar constant is unnecessar. Since x + 1 > 0 and ln a = ln a for real a > 0, this reduces to ln ((x + 1) 3 ). Therefore, our integrating factor is µ(x) = e ln ((x +1) 3 ) = (x + 1) 3. The next step is to multipl oth sides of our modified equation + ( 3x ) = x +1 integrating factor to simplif our work. 6x x +1 µ(x)( + ( 3x 6x )) = µ(x)( ) = x +1 x +1 (x + 1) 3 + (x + 1) 3 ( 3x ) = x +1 (x + 1) 3 ( 6x ). x +1 This simplifies down to (x + 1) 3 + 3x(x + 1) 1 = 6x(x + 1) 1. the Notice that d dx ((x +1) 3 ) = 3 (x +1) 1 (x +1) = 3 (x +1) 1 x = 3x(x +1) 1. This matches the second term in the equation. Therefore, we have: (x + 1) 3 + ((x + 1) 3 ) = 6x(x + 1) 1. Recalling the product rule, we know that fg + f g = (fg) with f = (x + 1) 3, g =, our equation simplifies down to: ((x + 1) 3 ) = 6x(x + 1) 1. We can now solve for integration. Appling the Fundamental Theorem of Calculus Part I, we know that d (f(x))dx = f(x) + C. dx Therefore: d dx ((x + 1) 3 )dx = 6x(x + 1) 1 dx = ((x + 1) 3 ) = 6x(x + 1) 1 dx and so = 6x(x +1) 1 dx (x +1) 3. To evaluate the integral in the numerator, we need to appl a u-sustitution. With u = x + 1, du = x = 1 du = xdx. This means that: dx
6x(x + 1) 1 dx = 6 (u) 1 1 du = 3 u 1 du = 3 3 u 3 + C = (x + 1) 3 + C since u n du = un+1 n+1 + C for real n 1. We can now write down our general solution: (t) = We are almost finished. 6x(x +1) 1 dx (x +1) 3 Appling our initial condition (0) = 1, we have: ((0) +1) 3 +C = 1+C 1 = 1 = C = 3. ((0) +1) 3 Our particular solution is then given : (t) = (x +1) 3 3 3 (t) = (x + 1) 3. (16 pts) Consider the initial value prolem (x +1) 3 = (x +1) 3 +C. (x +1) 3 = 1 or equivalentl: = 1 + cos(t), (0) = 0 (a) (6 pts) Solve using Integrating Factor Method hand. Our equation is in the form + p(t) = g(t), meaning we can use the Method of Integrating Factors directl. With p(t) = 1, g(t) = 1 + cos t we are read to compute our integrating factor. Our integrating factor is µ(t) = e p(t)dt = e 1dt = e t+c. Again, we can ignore the aritrar constant and take our integrating factor to e e t. The next step is to multipl oth sides of the equation this function: e t ( ) = e t (1 + cos t) = (e t ) + ( e t ) = e t + e t cos t. Noticing that (e t ) = e t like the previous prolem, we have an expression which reminds us of the product rule: (e t ) + (e t ) = e t + e t cos t = (e t ) = e t + e t cos t. We can now solve for : = (e t +e t cos t)dt e t. To tackle the integral aove, we can first split it up: (e t + e t cos t)dt = e t dt + e t cos tdt. The first integral can e tackled a u-sustitution with u = t = du = dt: e t dt = e u du = e u + C = e t + C. The second integral requires a recursive integration parts(ibp). It is a special case of the following integral: e at cos tdt. To solve an integral like this, we want to integrate parts until we see a similar integral appear. This will require appling IBP twice. The IBP formula is given udv = uv vdu. Setting u = e at, dv = cos t, du = (e at ) dt = ae at dt. v is the antiderivative of cos t = v = 1 sin t, appling the single-variale Chain Rule: d (f(g(t)) = f (g(t)) g (t) = d sin t cos t cos t ( ) = (t) = = cos t. dt dt
e at cos tdt = uv vdu = e t sin t ae at sin t dt = e t sin t a( e at sin tdt). Appling the same process with the new integral in parentheses: e at sin tdt = UdV = UV V du Setting U = e at, du = ae at dt, dv = sin t = V = Putting it all together: e at sin tdt = eat cos t ae at cos tdt = eat cos t + a e at cos tdt Simplifing, we have: e at cos tdt = eat sin t a( eat cos t + a e at cos tdt) = eat sin t + ( a e at cos t a e at cos tdt) cos t Continuing: e at cos tdt = eat sin t + a e at cos t a ( e at cos tdt) = (1 + a ) e at cos tdt = eat sin t + a e at cos t This simplifies to: a + ( e at cos tdt) = eat sin t + a e at cos t = e at cos tdt = ( eat sin t a + + a e at cos t) + C = eat (a cos t+ sin t) a + + C. The integral that we were tring to solve is a special case of the aove formula we derived. With a = 1, = 1, we have: e at cos tdt = eat (a cos t+ sin t) + C a + = e t cos tdt = e t ( cos 1 t+1 sin 1 t) + C = e t (sin t cos t) ( 1) +(1) Convenientl, e t cos tdt = e t (sin t cos t) + C. This means that our general solution is: (t) = (e t +e t cos t)dt e t = 1 + sin t cos t + C e t = e t +e t (sin t cos t)+c e t = sin t (1 + cos t) + Ce t Appling the initial condition (0) = 0, we otain: sin 0 (1 + cos 0) + Ce 0 = 0 = 0 + C = 0 = C = 0 +. The particular solution to the ODE must then e: (t) = sin t (1 + cos t) + ( 0 + )e t. () (4 pts) i. For what values of 0 does lim t (t) =? Recall that sin t 1 and cos t 1. This means that 1 sin t 1, 1 cos t 1. This implies that the expression sin t (1 + cos t) is at least 1 (1 + 1) = 3 and at most 1 (1 1) = 1 therefore 3 sin t (1 + cos t) 1, this term is ounded. Since the term ( 0 + )e t appears in the equation, as t this term will dominate: it will either vanish or grow quickl to ± depending on the value of 0. This is what we need to determine our long term ehavior. If 0 + > 0, lim t ( 0 + )e t = + = lim t (t) = + So we need 0 > to guarantee upward growth. ii. For what values of 0 does lim t (t) =? B the same reasoning, if ( 0 + ) < 0, lim t ( 0 + )e t = = lim t (t) =. So we need 0 < to guarantee oundless deca.
(c) (6 pts) Graph the direction field and 3 particular solutions on one graph using technolog (Maple is proal easiest). For the three particular solutions, choose 0 so that one solution goes to +, one goes to, and one remains ounded as t. Be sure to lael each curve. B our reasoning from part (), we need 0 > for +, 0 < to guarantee and we can pick 0 = to ensure that ( 0 +)e t = 0 so that (t) = sin t (1+cos t), which we showed must oscillate etween -3 and 1, and thus remains ounded aove and elow. The following Maple code generates a lue graph with 0 = 0 approaching +, a green graph with (0) = with confined oscillations and a purple graph with (0) = 3, approaching. with(detools): prolem1:=diff((t),t)=(1+*cos(t)) + (t); vars:={(t)}; t_values:=t=-6..6; _values:==-5..5; ICs:={[0,0],[0,-], [0,-3]}; DEplot(prolem1,vars,t_values,_values,ICs,stepsize=0.1,linecolor=[purple, green, lue]); 3. (10 pts) Consider the following initial value prolem d dx = + 1, (1) = 3. (a) Solve the initial value prolem using Separation of Variales. The Separation of Variales Technique allows us to solve first-order ODEs which can e written as a product of functions of the dependent and independent variales respectivel. We will assume x to e the independent variale. Setting h() = +1 d, g(x) = 1 we have: = ( +1 )(1) = (h())(g(x)), the equation is dx separale. We can then separate the functions and integrate: +1 = 1 = +1 dx = 1dx = d = 1dx +1 Since x, we know that from studing differentials and linear approximations that this relation carries through the approximation: d = dx.
To solve the integral on the left-hand side, we can use a u-sustitution with u = + 1, du = = d = 1 du, therefore: d d = 1 1du = 1 ln u + C = 1 ln +1 u + 1 + C Finishing up our solution: d = 1dx = 1 ln +1 + 1 = 1dx = x + C after comining constants. We can solve for. Notice that + 1 > 0 for all real-valued, which means we can drop the asolute value: ln + 1 = x + C = + 1 = Ce x = (x) = ± Ce x 1. Notice that we have two families of functions rather than one due to the ±. The real-valued function = x has a nonnegative range. This means that the sign indicated the initial condition tells us which solution to choose. Our initial condition is (1) = 3. This means that when x = 1, (x) = 3. Since the square root function does not change sign for real inputs, we can assume that the negative output of the solution (x) at one point implies a negative value for an other point in its range. Therefore, (x) = Ce x 1. We can now solve for C: 3 = Ce 1 1 = 9 = Ce 1 = C = 10. e The particular solution is given : 10 (x) = e e x 1 or equivalentl: (x) = 10e x 1. () For what values of x, does the solution exist? (This means the solution is well-defined and is well-defined). Appling the Chain Rule, the derivative is given : = d ( 10e dx x 1) = 1 (10ex 1) 1 (10e x 1) = 1 (10ex 1) 1 (10(x ) e x 0) = 1 (10ex 1) 1 (10()e x 0) = (10e x 1) 1 (10e x ) Finall, = 10e x 1 = = 10ex. 10e x 1 The derivative does not exist when 10e x 1 0 and that is enough to guarantee discontinuit in our solution. The solution exists when: 10e x 1 > 0 e x > e x > 1 e ln x > 1 ( ln e ln 10) 10 10 This means that ln 10 x > 1 0.1519 (c) Graph the solution and denote on the graph the interval of x-values otained from part () where the solution exists. Explain graphicall wh it does not exist eond that interval. The following Maple code generates a direction field for the ODE and a graph of the ln 10 solution in lue. The lue graph nearl touches the x-axis at x = 1 : a warning is given maple claiming that the solution cannot e evaluated eond this point.
with(detools): prolem1:=diff((x),x)=(((x))^ + 1)/((x)); vars:={(x)}; x_values:=x=-6..6; _values:==-5..5; ICs:={[1,-3]}; DEplot(prolem1,vars,x_values,_values,ICs,stepsize=0.1,linecolor=[purple]); Correspondingl, the graph is not visile for x-values to the left of this point. That is ecause near this point, the derivative tends to infinit, as shown the vertical tendenc of the graph near the x-intercept. ln 10 Furthermore, since the graph is complex-valued for x < 1, it the graph will not e visile in this region, as we can see. Producing this graph using other software such as Desmos, our output looks like the following: The point (1 ln 10, 0) is emphasized.