MATH 308, Spring 2016
QUIZ # 2
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SECTION #:
INSTRUCTOR: Dr. Marco A. Roque Sol
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1. (10 pts ) (Problem 13. Secc. 2.2) Solve the Initial Value Problem (IVP)
y0 =
.
Solution
2x
;
y + x2 y
y(0) = −2
2x
2x
dy
= y0 =
=
dx
y + x2 y
y(1 + x2 )
Separating and integrating the above equation
ydy =
2x
dx =⇒
(1 + x2 )
Z
Z
ydy =
2x
dx
(1 + x2 )
we obtain
y2
= ln|1 + x2 | + c = ln(1 + x2 ) + c
2
Applying initial conditions
(−2)2
= ln|1 + (0)2 | + c =⇒ 2 = ln((1) + c =⇒ c = 2
2
and the implicit solution is
y2
= ln|1 + x2 | + 2
2
Considering the initial condition, the explicit solution is
p
y = − 2ln|1 + x2 | + 4
2. (10 pts.)(Problem 25. Secc. 2.1) Solve the Initial Value Problem (IVP)
ty 0 − 2y =
.
sin(t)
;
t
y(−π/2) = a;
t<0
How does the solutions behave when t → 0 ? Does the behavior depend on the choice of the initial value a ?
Solution
2
sin(t)
sin(t)
=⇒ y 0 + y =
t
t
t2
ty 0 + 2y =
.
The integrating factor is giving by
µ(t) = e
R
p(t)dt
=e
R
2
t dt
= e2ln|t| = t2
.
multiplying the ODE by µ(t) and integrating, we have
2 0
2
Z
0
t y + 2ty = sin(t) =⇒ (t y) = sin(t) =⇒
2
0
(t y) dt =
.
=⇒ t2 y = −cos(t) + c =⇒ y = −
.
cos(t)
c
+ 2
2
t
t
And applying initial conditions
a=−
.
cos(−π/2)
c
aπ 2
+
=⇒ c =
2
2
(−π/2)
(−π/2)
4
we obtain the particular solution
y=−
.
Now if a =
4
π2
then
y=−
.
cos(t) aπ 2
+ 2
t2
4t
cos(t)
1
1 − cost
+ 2 =
2
t
t
t2
and
lim y =
.
t→0
1
2
3. (10 pts.) (Problem 19. Secc. 2.2) Solve the Initial Value Problem (IVP)
sin(2x)dx + cos(3y)dy = 0;
Solution
y(π/2) = π/3
Z
sin(t)dt
The ODE is separable, therefore
Z
Z
sin(2x)dx = −cos(3y)dy =⇒
sin(2x)dx = −
.
1
1
cos(3y)dy =⇒ − cos(2x) = −
2
3
Z
sin(3y) + c
Appliying initial conditions
1
1
1
− cos(2(pi/2)) = − sin(3(π/3)) + c =⇒ c =
2
3
2
.
Therefore the implicit solution is
cos(2y) =
.
2
sin(3x) − 1
3
and the explicit solution is
1
y = cos−1
2
.
2
sin(3x) − 1 ;
3
−1 ≤
2
sin(3x) − 1 ≤ 1
3
4. (10 pts.)(Bonus) (Problem 38. Secc. 2.1) Using the method of Variation of Parameters. Assume that the
solution of
y 0 + p(t)y = g(t)
.
is of the form
y(t) = A(t)e−
R
p(t)dt
.
and nd a ODE satised by A(t)
Solution
If the solution is of the form
y(t) = A(t)e−
R
p(t)dt
=⇒ y(t) = A0 (t)e−
R
p(t)dt
R
0
R
R
+ A(t) e− p(t)dt = A0 (t)e− p(t)dt − A(t)p(t)e− p(t)dt
.
=⇒ y(t) = A0 (t)e−
R
p(t)dt
− p(t)y(t) =⇒ y(t) + p(t)y(t) = A0 (t)e−
.
but we have that y(t) + p(t)y(t) = g(t), so
A0 (t)e−
.
therefore
R
p(t)dt
= g(t)
R
p(t)dt
A0 (t) = e
R
p(t)dt
g(t) = h(t)g(t);
h(t) = e
.
Thus, we get
Z
A(t) =
.
h(t)g(t)dt (c =??)
R
p(t)dt