Ordinary Differential Equations
MATH 308 - 523
A. Bonito
February 17
Spring 2015
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Quiz 2
— 5 minute individual quiz;
— Answer the questions in the space provided. If you run out of space, continue
onto the back of the page. Additional space is provided at the end;
— Show and explain all work;
— Underline the answer of each steps;
— The use of books, personal notes, calculator, cellphone, laptop, and communication with others is forbidden;
— By taking this quiz, you agree to follow the university’s code of academic
integrity.
Exercise 1
20%
Discuss the existence and uniqueness of a solution to the following ordinary differential equation
(JUSTIFY your argumentation)
(1 + x2 )y 00 (x) + xy 0 (x) − y(x) = ex ,
y(1) = 1, y 0 (1) = 1.
Exercise 2
80%
Solve the following initial value problem :
2y 00 + 7y 0 − 4y = 0,
y(0) = 9/4,
y 0 (0) = 0.
Ordinary Differential Equations
MATH 308 - 523
A. Bonito
February 17
Spring 2015
Quiz 2: solutions
Exercise 1
20%
First we rewrite the ODE to fit the existence/uniqueness theorem framework dividing by 1 + x2
y 00 (x) +
1
ex
x
y 0 (x) −
y(x) =
.
2
2
1+x
1+x
1 + x2
| {z }
| {z }
| {z }
p(x)
q(x)
g(x)
Since p(x), q(x) and g(x) are continuous for all x ∈ R, there exists a unique solution y(x) for all
x ∈ R.
Exercise 2
80%
Guessing y(x) = eλx we obtain the characteristic equation
2λ2 + 7λ − 4 = 0.
Two find the roots of the above polynomial, we compute the discriminant
∆ = 72 − 4 × 2 × (−4) = 81 = 92
so that the two distinct roots are given by
λ1 = 1/2
and
λ2 = −4.
Therefore, two linearly independent solutions read
y1 (x) = ex/2
and
y2 (x) = e−4x .
The general solution is given by
y(x) = C1 ex/2 + C2 e−4x
where C1 , C2 are constants to be determined using the initial conditions :
C1 + C2 =
9
4
and
1
C1 − 4C2 = 0.
2
Solving the above system leads to
C1 = 2
and
C2 =
so that the desired solution is given by
1
y(x) = 2ex/2 + e−4x .
4
1
4