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MATH 1B Lecture 15: Math 1B - Calculus - Lecture 15_Finishing Up Differential Equations

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26 Mar 2015

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Professor Coward

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Math 1b: calculus - lecture 15: finishing up differential equations y^2 (dy/dx) dx = 2x + 3 (x) dx. A: we have 2yey^2 (dy/dx) = 2x + 3 (x). 2ye e y^2 = x2 + 2 x3/2 + c, where c is a constant. y 2 = ln |x2 + 2x3/2 + c|. A: we have x(dy/dx) - y = xln(x). (dy/dx) - (y/x) = ln(x). Convert it to the form (dy/dx) + p(x)y = q(x). E-ln|x| = e-ln(x) = eln(x^-1) = 1/x. Remember the integrating factor is e p(x) dx. 2)y = ln(y) / x. (1/x)(dy/dx) - (1/x (dy/dx) [(1/x) y] = ln(x) / x. (y/x) = [ln(x) / x] dx. (y/x) = ln(x) y = (x/2) ln(x) 2 + c, where c is a constant. So 2/1 = c. c = 2. y = x[(ln(x) Use euler"s method with a step size of . 5 to estimate y(1). A: we have xo = 0; y0 = 1.

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Calculus

ISBN: 9781319050733

Single Variable Calculus: Early Transcendentals

4th Edition, 2018

ISBN: 9781337687805

CALCULUS:EARLY TRANSCENDENTALS

ISBN: 9781319050740

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Related Questions

1.Solve the problem. Find the demand function q = D(x), given that E(x) = 8/x and q = e when x=8 a) q = e^8/x b) q = ln 8/x c) q = 8 lnx d) q = ln^8/x

2. Solve 3y^2 dy/dx = 7x a) y = - 3sqrt(7/2 x^2 +C) b) y = - 3sqrt(7x +C) c) y = 3sqrt(7x^2 +C) d) y = 3sqrt(7/2 x^2 +C)

3. Find the general solution for the differential equation.
y ' = 72x^2 - 20x a) 72x^3 - 20x^2 + C b) 72x^3 - 10x^2 + C c) 24x^3 -10x^2 + C d) 24x^3 - 20x^2 + C

Follow the six steps below to solve the following: xy' + 6xy =xe^(5x)

Step 1: REWRITE into standard form (y' + P(x)y + Q(x))

a) How does the given equation differ from standard form?

b) What can be done to put the given equation into standardform?

Step 2: IDENTIFY P(x) and Q(x).

a) Where does the standard form indicate P(x) and Q(x) arelocated?

P(x) =

Q(x) =

Step 3: DETERMINE the integrating factor (e^(?P(x) dx)). PlugP(x) as identified in step 2 into the integrating factor forme^(?P(x) dx) . Hint: Simplify e^(?P(x) dx) if possible

a) What is meant by the integrating factor?

b) If the integrating factor looks like e^(ln (x)) how could itbe simplified?

c) What about e^(1/2 ln (x))?

d) What about e^(-ln (x))?

?P(x) dx =

e^(?P(x) dx) =

Step 4: SUBSTITUTE Q(x) from step 2 and e^(?P(x) dx) from step 3into the solution form ye^(?P(x) dx) = ?Q(x)e^(?P(x) dx) + C.

Step 5: INTEGRATE the right side of the result of step 4. Hint:You may need to simplify and/or rewrite the right side before youattempt to integrate.

Step 6: SOLVE for y. To do this divide both sides by theintegrating factor. Remember that all parts of the right sideincluding the C must be divided by the integrating factor.

Solution is _________________________

help

Using the function y = f(x) obtained in (c), find the .y coordinate corresponding to the value x = -1 / 2 Evaluate both formulas for dy / dx at this point. Why do you think this happens.? Using the explicit derivative formula dy / dx = f (x) obtained in part (c) , find all points on the graph of y = f(x) where the tangent line is horizontal. Then substitute the x and y - coordinates of each such point into the implicit dy / dx formula given in part(b). Again, why does this happen? Again using the explicit derivative formula dy / dx = f(x) from part (c). find all points on the graph of f where the tangent slope equals 6. Now substitute the x- and y- coordinates of each such point into the implicit dy / dx formula given in part (b) to confirm this slope value. It is clear that no point (x, y) for which x + 2y = 0 will appear on the graph of original equation 2x = y - x2 / x +2y. In effect, points on the line x + 2y = 0 are "cut out" of the graph of the function f Relate this observation to the phenomena encountered earlier in parts (e) and (f). Once again referring to the explicit function f(x) obtained in part (c), where does the graph of f have a vertical asymptote? What happens when you substitute this X-value into the original implicit equation (given at the beginning of this worksheet) and attempt to solve for y ? Explain. [ Verification of the implicit derivative given in part (b).] Implicitly differentiate the equation 2x = y - x2 / x = 2y, applying the quotient role on the right-hand side. Clear the denominator and expand terms, then solve for dy / dx and simplify. Note: Solutions to this problem set are t0 be established by traditional algebraic or arithmetic means. using Maple software to verify each result To download the Maple template file prepared to assist you w ith these exercises, go to https;//my,ltu edu and login, select Calculus 1 and click on Workshop. Saw this file to your laptop hard drive and exit the web Hand-generated solutions are to be fully documented In the workshop report including all steps requiring algebraic or calculus computations - merely listing the answers produce with Maple is not sufficient. Printouts of your Maple verifications are required as part of your workshop report. Consider the equation 2x = y -x2 / x+ 2y. Verify that the points .A(1, -1) and B(-2, 4/3) lie on the graph of this equation. In part (j) of this worksheet you will be asked to verify- the derivative formula by implicitly differentiating the equation given above. Evaluate this derivative at points A and B. Solve for y in the original equation, expressing y as an explicit function of x. (For later reference, we will call this function f(x).) Compute dy / dx from this expression and simplify , then evaluate this explicit version of dy / dx at both A aud B. Considering the results of part (b), are these the values you expected? Why or why not? Evaluate each of the two formulas for dy / dx (both implicit and explicit) at the point C(-1, 0). Can you explain your results?