Consider8 the homogenous linear second order differentialequation
y′′ −10y′ +25y = 0 (⋆)
Follow the presentation of Example 2 in the handwritten notesfor class 15 on March 24 closely to eventually find the generalsolution to this differential equation. That is:
(a) The first key idea is the same as in our solution ofquestion 2, namely we assume that a solution y has the form
.
y = erx for some constant real number r
′ ′′
(b) Using the key idea from part a), compute the derivatives yand the y .
(c) Substitute the expressions for y, y′ and y′′ that youobtained in parts a) and b) into the differential equation (⋆);this leads to an equation that must necessarily hold for the realnumber r. Dividing this equation by the expression erx which cannever be equal to zero, will then lead to a quadratic equation thatmust necessarily hold for the constant number r. Find thisquadratic equation for the number r.
Remark. This quadratic equation for the variable r is calledthe characteristic equation of the differential equation (⋆).
(d) Solve the characteristic equation9 from part c) in orderto find its unique real solution r.
8This question is closely related to Example 2 on pages 6-9 ofour handwritten class notes for Class 15 on March 24; it is alsoclosely related to Example 5.2.2 on page 212 of our textbook.
9If you work correctly, you will obtain the characteristicequation r2 − 10r + 25 = 0, whose only existing solution is r =5.
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(e) Since we only obtained a single solution in part d), weneed a second key idea in order to find a second, linearlyindependent,10 solution: Namely, now that we have one solution ofthe form y = e5x, we assume furthermore that a solution y has theform
y = u · e5x for some unknown function u = u(x) that depends onthe variable x.
(f) Use the product rules and the assumption from part e) tocarefully compute the derivatives y′ and
′′ 5x 11 the y . In doing so, make sure to factor theexpression e in each expression for the derivatives.
(g) Substitute the expressions for y, y′ and y′′ that youobtained in parts e) and f) into the differential equation (⋆);this will lead to a second order differential equation that mustnecessarily hold for the function u = u(x). Dividing this equationby the expression e5x which can never be equal to zero, this willto lead to an easy differential equation for u. Find12 thisdifferential equation for the function u.
(h) Solve the differential equation13 that you obtained inpart g) by taking antiderivatives twice. Make sure to not forgetthe two different constants of integration that arise in thisprocess, and let’s agree to call them c and d.
10Two solutions f(x) and g(x) are called linearly independent,if f(x) and g(x) are not constant multiples of each other. In otherwords, f(x) and g(x) are linearly independent, if the quotientfunction f(x) is not a constant function.
g(x)
11If you work correctly, you will compute that y′ = e5x · (u′+ 5u) and that y′′ = e5x · (u′′ + 10u′ + 25u).
12If you work correctly, you will obtain the differentialequation u′′ = 0.
13If you work correctly, you will take antiderivatives twiceto solve the equation u′′ = 0 and obtain that u′ = c andconsequently that u = cx+d for any constants c and d.
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(i) Using the assumption from part e) and your solution fromh) state your solution function y to the differential equation(⋆).
(j) Use two different pairs of constants c and d to find twodifferent solutions f(x) and g(x) with the
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goal of showing that the two functions will turn out linearlyindependent solutions • one solution function f(x) corresponding tothe pair c = 0 and d = 1:
. That is, find:
• one solution function g(x) corresponding to the pair c = 1and d = 0:
(k) Show that the two solutions that you obtained in part j)are linearly independent
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(l) Use your work in parts a)- k) to find the generalsolution16 to the differential equation (⋆). Express your answerusing constants c and d.
