MATH1251 Calculus Test 1 2010 S1 v1A
Full Solutions
August 23, 2015
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1. To find this integral, we must decompose it into partial fractions as follows,
4x2
A
Bx + C
≡
+ 2
.
2
(x − 2) (x + 4)
x−2
x +4
Applying the Heaviside Cover-up method, we can easily find A,
A=
4 (2)2
= 2.
(2)2 + 4
To solve for B and C, we first multiply both sides by (x − 2) x2 + 4 to obtain,
4x2 ≡ 2 x2 + 4 + (Bx + C) (x − 2) .
Equating the coefficient of x2 , we find that 4 = 2 + B which implies that B = 2. (A tip for
1
applying this method generally in finding coefficients is to work your way down in terms
of powers!). Next, equating the constants, we find that 0 = 8 − 2C and thus C = 4. Our
integral now becomes something that can easily be solved,
Z
Z
4x2
2
2x + 4
dx =
+
dx
(x − 2) (x2 + 4)
x − 2 x2 + 4
x
= 2 ln |x − 2| + ln x2 + 4 + 2 tan−1
+ c.
2
2. To find this recurrence relation, we notice that in the required expression, we have the
power of x decrease by one. This implies that we should integrate ekx and differentiate
xn when applying integration by parts.
1
Z
xn ekx dx
In =
0
xn ekx
=
k
1
Z
−
0
1
nxn−1
0
ekx
dx
k
Z
n 1 n−1 kx
ek
x
e dx
−
=
k
k 0
ek
n
=
− In−1 ,
k
k
as required.
3. We notice that the equation is in the form
dy
+ f (x)g(y) = h(x).
dx
This means that we can solve this differential equation using an Integration Factor, i.e.
R
I(x) = exp f (x) dx .
R
So, the integration factor for this particular question is I(x) = exp − x2 dx = x12 .
Multiplying this through the equation, we obtain
1 dy
2
− y = 6x2 .
x2 dx x3
Applying this method essentially creates a product rule differentiated expression on the
left, so reversing this,
d y
= 6x2 .
dx x2
Integrating both sides and simplifying, we obtain our answer,
y
= 2x3 + c
x2
∴ y = 2x5 + cx2 ,
2
where c ∈ R.
4.
9x2 y 2 dx + 6x3 y + 3y 2 dy = 0.
First, let u (x, y) = 9x2 y 2 and v (x, y) = 6x3 y + 3y 2 .
To show that this differential equation is exact, we must show that
Clearly,
∂u
∂y
= 18x2 y and
∂v
∂x
= 18x2 y, and thus
∂u
∂y
=
∂v
∂x
∂u
∂y
=
∂v
∂x .
implying that the differential
equation is exact.
To solve this, we let our solution be H (x, y) which must satisfy
∂H
∂x
= u and
∂H
∂y
= v.
Taking the first equation (it doesn’t matter which one), we integrate both sides with
respect to x,
∂H
= 9x2 y 2
∂x
=⇒ H (x, y) = 3x3 y 2 + c (y) .
Note that a term dependent on y pops up because such a term would have been eliminated
due to the partial differentiate with respect to x.
Now partially differentiating with respect to y,
∂H
= 6x3 y + c0 (y) .
∂y
Comparing this to what
∂H
∂y
should be equal to, as it should be equal to v (x, y),
6x3 y + c0 (y) ≡ 6x3 y + 3y 2 .
This implies that c0 (y) = 3y 2 , and hence c (y) = y 3 + c, where c is a constant.
Thus, the solution is,
H (x, y) = 3x3 y 2 + y 3 + c = 0,
which is equivalent to (and the answer is more commonly written as..),
3x3 y 2 + y 3 = c.
3