Question

Consider8 the homogenous linear second order differential equation

y′′ −10y′ +25y = 0 (⋆)

Follow the presentation of Example 2 in the handwritten notes for class 15 on March 24 closely to eventually find the general solution to this differential equation. That is:

(a) The first key idea is the same as in our solution of question 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 y and the y .

(c) Substitute the expressions for y, y′ and y′′ that you obtained in parts a) and b) into the differential equation (⋆); this leads to an equation that must necessarily hold for the real number r. Dividing this equation by the expression erx which can never be equal to zero, will then lead to a quadratic equation that must necessarily hold for the constant number r. Find this quadratic equation for the number r.

Remark. This quadratic equation for the variable r is called the characteristic equation of the differential equation (⋆).

(d) Solve the characteristic equation9 from part c) in order to find its unique real solution r.

8This question is closely related to Example 2 on pages 6-9 of our handwritten class notes for Class 15 on March 24; it is also closely related to Example 5.2.2 on page 212 of our textbook.

9If you work correctly, you will obtain the characteristic equation 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), we need a second key idea in order to find a second, linearly independent,10 solution: Namely, now that we have one solution of the form y = e5x, we assume furthermore that a solution y has the form

y = u · e5x for some unknown function u = u(x) that depends on the variable x.

(f) Use the product rules and the assumption from part e) to carefully compute the derivatives y′ and

′′ 5x 11 the y . In doing so, make sure to factor the expression e in each expression for the derivatives.

(g) Substitute the expressions for y, y′ and y′′ that you obtained in parts e) and f) into the differential equation (⋆); this will lead to a second order differential equation that must necessarily hold for the function u = u(x). Dividing this equation by the expression e5x which can never be equal to zero, this will to lead to an easy differential equation for u. Find12 this differential equation for the function u.

(h) Solve the differential equation13 that you obtained in part g) by taking antiderivatives twice. Make sure to not forget the two different constants of integration that arise in this process, 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 other words, f(x) and g(x) are linearly independent, if the quotient function 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 differential equation u′′ = 0.

13If you work correctly, you will take antiderivatives twice to solve the equation u′′ = 0 and obtain that u′ = c and consequently that u = cx+d for any constants c and d.

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(i) Using the assumption from part e) and your solution from h) state your solution function y to the differential equation (⋆).

(j) Use two different pairs of constants c and d to find two different solutions f(x) and g(x) with the

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goal of showing that the two functions will turn out linearly independent solutions • one solution function f(x) corresponding to the pair c = 0 and d = 1:

. That is, find:

• one solution function g(x) corresponding to the pair c = 1 and 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 general solution16 to the differential equation (⋆). Express your answer using constants c and d.

Answer 1