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Assignment 1 (MATH 215, Q1)
1. Solve the differential equation.
(a) y 0 = y 2 sin x
Solution. We have
dy
= sin x dx.
y2
Integrating both sides gives
−
1
= − cos x + C.
y
Hence, the general solution of the differential equations is
y=
dy
2y
=
.
dx
x
Solution. We have
1
.
cos x − C
(b)
2dx
dy
=
.
y
x
Integrating both sides gives
or |y| = e2 ln |x|+C1 .
ln |y| = 2 ln |x| + C1
Hence, the general solution of the differential equations is
y = Cx2 .
2. Solve the given first-order linear equation and verify that your solution indeed satisfies
the equation.
2
(a) y 0 − 2xy = 2xex
R
2
− 2x dx
Solution. An integrating factor is I(x) = e
= e−x . Multiplying both sides by
2
e−x gives
2
2
d −x2 e−x y 0 − 2xe−x y = 2x or
e
y = 2x.
dx
R
2
It follows that e−x y = 2x dx = x2 + C. Therefore, the general solution is
2
2
y = x2 ex + Cex .
1
2
2
Let us verify that y = x2 ex + Cex is the general solution of the differential equation.
We have
2
2
2
y 0 = 2xex + x2 (2x)ex + C(2x)ex .
Hence,
2
2
2
2
2
2
y 0 − 2xy = 2xex + x2 (2x)ex + C(2x)ex − (2x)x2 ex − (2x)Cex = 2xex .
dy
y
1
= + 2 , x > 0.
dx
x x
Solution. Rewrite the equation in the form
(b)
−
I(x) = e
R
dx
x
dy
dx
y
x
=
1
x2 .
= e− ln x =
1
.
x
−
An integrating factor is
Multiplying both sides by 1/x gives
1 dy
y
1
− 2 = 3
x dx x
x
It follows that
y
=
x
Z
or
d y
1
= 3.
dx x
x
1
1
dx = − 2 + C.
3
x
2x
Therefore, the general solution is
y = Cx −
1
,
2x
x > 0.
For verification we observe that
left =
1
dy
=C+ 2
dx
2x
and right =
Since left = right, y = Cx −
1
2x
y
1
1
1
1
+ 2 = C − 2 + 2 = C + 2.
x x
2x
x
2x
is indeed the general solution.
3. Solve the initial-value problem.
(a) y 0 + y = x + ex , y(0) = 0
Solution. An integrating factor is I(x) = ex . Multiplying both sides by ex gives
ex y 0 + ex y = ex (x + ex )
or
d x
(e y) = xex + e2x .
dx
It follows that
x
e y=
Z
1
xex + e2x dx = (x − 1)ex + e2x + C.
2
2
So the general solution is
y =x−1+
1 x
e + C e−x .
2
Set x = 0 and y = 0 in the above equation: 0 = −1 +
therefore the desired solution is
1
2
+ C. We obtain C = 12 , and
1
1
y = x − 1 + ex + e−x .
2
2
dy
+ 2y = t3 , t > 0, y(1) = 0
dt
Solution. The equation can be rewritten as
(b) t
dy 2
+ y = t2 , t > 0.
dt
t
R2
An integrating factor is I(t) = e t dt = e2 ln t = t2 . Multiplying both sides of the
equation by t2 , we get
t2
dy
+ 2ty = t4
dt
It follows that
2
Z
t y=
or
t4 dt =
d 2
(t y) = t4 .
dt
1 5
t + C.
5
Hence, the general solution of the equation is
y=
1 3 C
t + 2.
5
t
Set t = 1 and y = 0 in the above equation: 0 =
therefore the desired solution is
1
5
+ C. We obtain C = − 51 , and
1
t3
− 2.
y=
5
5t
4. Find the general solution of the given differential equation.
(a) y 00 − 4y 0 + 4y = 0
Solution. The characteristic equation is r2 − 4r + 4 = 0. It has a double real root
r = 2. The general solution of the equation is
y = C1 e2x + C2 xe2x .
3
(b) y 00 + 6y 0 + 10y = 0
Solution. The characteristic equation is r2 + 6r + 10 = 0. Its two roots are
r1,2 =
−6 +
√
62 − 4 · 10
2
or
r1 = −3 + i and r2 = −3 − i.
The general solution of the equation is
y = e−3x (C1 cos x + C2 sin x).
5. Solve the initial-value problem.
(a) y 00 + 3y 0 − 4y = 0, y(0) = 2, y 0 (0) = −3
Solution. The characteristic equation is r2 + 3r − 4 = (r + 4)(r − 1) = 0. The
general solution of the differential equation is y = C1 e−4x + C2 ex . It follows that
y 0 = −4C1 e−4x + C2 ex . The initial value conditions y(0) = 2 and y 0 (0) = −3 give
C1 + C2 = 2 and −4C1 + C2 = −3. Solving these two equations for C1 and C2 , we
obtain C1 = 1 and C2 = 1. Therefore, the desired solution is
y = e−4x + ex .
(b) y 00 + 4y = 0, y(π/6) = 1, y 0 (π/6) = 0
Solution. The characteristic equation r2 + 4 = 0 has two roots r1 = 2i and r2 = −2i.
The general solution of the differential equation is y = C1 cos(2x) + C2 sin(2x). It
follows that y 0 = −2C1 sin(2x) + 2C2 cos(2x). The initial value conditions y(π/6) = 1
√
√
and y 0 (π/6) = 0 give 21 C1 + 23 C2 = 1 and − 3C1 + C2 = 0. Solving these two
√
equations for C1 and C2 , we obtain C1 = 21 and C2 = 23 . Therefore, the desired
solution is
√
1
3
y = cos(2x) +
sin(2x).
2
2
4