Math Repository
about
topic
al
year
ly
Yearly
2008
P1 Q4
Topical
DE
08 P1 Q4
2008 H2 Mathematics Paper 1 Question 4
Differential Equations (DEs)
Answers
(i)
y
=
3
2
ln
(
x
2
+
1
)
+
c
{y = \frac{3}{2} \ln \left( x^2 + 1 \right) + c}
y
=
2
3
ln
(
x
2
+
1
)
+
c
(ii)
y
=
3
2
ln
(
x
2
+
1
)
+
2
{y = \frac{3}{2} \ln \left( x^2 + 1 \right) + 2}
y
=
2
3
ln
(
x
2
+
1
)
+
2
(iii)
The gradient of every solution curve approaches
0
{0}
0
as
x
→
±
∞
{x \to \pm \infty}
x
→
±
∞
(iv)
Graph of particular solution
Full solutions
(i)
d
y
d
x
=
3
x
x
2
+
1
y
=
3
2
∫
2
x
x
2
+
1
d
x
=
3
2
ln
(
x
2
+
1
)
+
c
■
\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} &= \frac{3x}{x^2 + 1} \\ y &= \frac{3}{2} \int \frac{2x}{x^2 + 1} \; \mathrm{d}x \\ &= \frac{3}{2} \ln \left( x^2 + 1 \right) + c \; \blacksquare \end{align*}
d
x
d
y
y
=
x
2
+
1
3
x
=
2
3
∫
x
2
+
1
2
x
d
x
=
2
3
ln
(
x
2
+
1
)
+
c
■
(ii)
When
x
=
0
,
{x=0, }
x
=
0
,
y
=
2
,
{y = 2,}
y
=
2
,
2
=
3
2
ln
(
0
2
+
1
)
+
c
c
=
2
\begin{gather*} 2 = \frac{3}{2} \ln (0^2+1) + c \\ c = 2 \end{gather*}
2
=
2
3
ln
(
0
2
+
1
)
+
c
c
=
2
Particular solution:
y
=
3
2
ln
(
x
2
+
1
)
+
2
■
y = \frac{3}{2} \ln \left( x^2 + 1 \right) + 2 \; \blacksquare
y
=
2
3
ln
(
x
2
+
1
)
+
2
■
(iii)
As
x
→
±
∞
,
{x \to \pm \infty, }
x
→
±
∞
,
d
y
d
x
=
3
x
x
2
+
1
→
0
\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{3x}{x^2 + 1} \to 0
d
x
d
y
=
x
2
+
1
3
x
→
0
Hence the gradient of every solution curve approaches
0
{0}
0
as
x
→
±
∞
■
{x \to \pm \infty \; \blacksquare}
x
→
±
∞
■
(iv)
Graph of particular solution
Back to top ▲