Handout 1
MATH 205, Fall 2020
Q1: What is the order of the DE xy (3) + y 2 = sin x? Is this a linear differential equation?
Q2: Let
an (x)y (n) + an−1 (x)y (n−1) + · · · + a1 (x)y 0 + a0 (x)y = 0
be a (homogeneous) linear equation. Show that if y1 and y2 are two solutions of this DE, then
y1 + y2 is also a solution. Is ay1 + by2 a solution (with a, b scalars)?
Q3: Is y =
x2
1
+ a solution to the DE xy 0 + y = x2 ? on which intervals?
3
x
Q4: Find m so that y = emx is a solution of y 00 − 5y 0 + 6y = 0.
Q5: Check that the implicit function y defined by sin y + xy − x3 = 2 is solution to the DE
d2 y
6xy 0 + (y 0 )3 sin y − 2(y 0 )2
=
dx2
3x2 − y
Q6: Explain why the IVP
√
0
y =
y−2
, y(1) = 1
x
has no solution.
Q7: Is the function
(
0, x < 0
y=
x, x ≥ 0
a solution of the IVP: xy 0 = y, y(0) = 0.
Q8: (existence and uniqueness) Find a point (x0 , y0 ) so that the given DE: (y − x)y 0 = y + x with
initial condition y(x0 ) = y0 has a unique solution on some interval around x0 .
Q9: Problem 47,§1.2 in the book.
1
2
Solutions Handout 1
Q1: Order = 3. No.
Q2: Yes. Plug in and check.
Q3: Yes...check by direct differentiation. This is a solution on either (0, ∞) or on (−∞, 0). y isn’t a
solution of the differential equation on any open interval that contains x = 0, since y is not defined
at x = 0.
Q4: Differentiate and replace you get two possible values; m = 2 or m = 3 (check).
Q5: You differentiate sin y + xy − x3 = 2 with respect to x, you get y 0 cos y + y + xy 0 − 3x2 = 0.
You differentiate a second time, you get
y 00 cos y − (y 0 )2 sin y + y 0 + y 0 + xy 00 − 6x = 0
Multiply by y 0 and replace y 0 cos y by 3x2 − y − xy 0 . Collect terms in y 00 you get the answer.
This exercise to train you on implicit differentiation.
Q6: For this IVP to have a solution y, the function y must satisfy y(x) − 2 ≥ 0 for all x in its
interval of existence. Obviously y(1) = 1 is an impossible choice.
Q7: With this choice of y it is easy to verify that the DE is satisfied on every interval not containing
0 and xy 0 = y. Moreover y(0) = 0 by definition. So it seems the answer is yes... BUT let’s not
forget that a solution for the first order DE must be continuous and differentiable on its interval of
existence. The function y given above is continuous but is not differentiable at x = 0 (why? check
that the slope as a limit of a quotient on the left of 0 is 0, while on the right of 0 it is 1). It is not
a solution.
Q8: Write this DE as
y+x
y−x
We notice that the function of two variables f (x, y) =
y0 =
y+x
y−x
is defined and continuous in the region
of the (xy)-plane where y 6= x. It has a continuous first partial derivative ∂f
∂y in that region as well
since
∂f
−2x
=
∂y
(y − x)2
By Theorem 1.2.1 in the book (on existence of unique solution), it suffices to choose a point (x0 , y0 )
so that both f and ∂f
∂y are continuous at that point. Can choose any point (x0 , y0 ) with x0 6= y0 ...
Q9: At (0, 21 ), y 0 (0) = x − 2y = 0 − 1 = −1 and the curve must have negative slope at that point.
Select the red curve.