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18 P1 Q3
2018 H2 Mathematics Paper 1 Question 3
Differential Equations (DEs)
Answers
(i)
d
u
d
x
=
−
6
x
3
{\displaystyle \frac{\mathrm{d}u}{\mathrm{d}x} = -\frac{6}{x^3}}
d
x
d
u
=
−
x
3
6
(ii)
y
=
3
−
x
2
{y=3 - x^2}
y
=
3
−
x
2
Full solutions
(i)
y
=
u
x
2
d
y
d
x
=
d
u
d
x
x
2
+
2
u
x
\begin{gather*} y = ux^2 \\ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\mathrm{d}u}{\mathrm{d}x}x^2 + 2ux \\ \end{gather*}
y
=
u
x
2
d
x
d
y
=
d
x
d
u
x
2
+
2
ux
x
d
y
d
x
=
d
u
d
x
x
3
+
2
u
x
2
\begin{equation} \quad x\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\mathrm{d}u}{\mathrm{d}x}x^3 + 2ux^2 \end{equation}
x
d
x
d
y
=
d
x
d
u
x
3
+
2
u
x
2
Substituting
y
=
u
x
2
{y=ux^2}
y
=
u
x
2
and
(
1
)
{(1)}
(
1
)
into the original differential equation,
d
u
d
x
x
3
+
2
u
x
2
=
2
u
x
2
−
6
d
u
d
x
x
3
=
−
6
d
u
d
x
=
−
6
x
3
■
\begin{gather*} \frac{\mathrm{d}u}{\mathrm{d}x}x^3 + 2ux^2 = 2ux^2 - 6 \\ \frac{\mathrm{d}u}{\mathrm{d}x}x^3 = -6 \\ \frac{\mathrm{d}u}{\mathrm{d}x} = -\frac{6}{x^3} \; \blacksquare \end{gather*}
d
x
d
u
x
3
+
2
u
x
2
=
2
u
x
2
−
6
d
x
d
u
x
3
=
−
6
d
x
d
u
=
−
x
3
6
■
(ii)
d
u
d
x
=
−
6
x
3
u
=
3
x
2
+
c
y
x
2
=
3
x
2
+
c
y
=
3
+
c
x
2
\begin{align*} \frac{\mathrm{d}u}{\mathrm{d}x} &= - \frac{6}{x^{3}} \\ u &= \frac{3}{x^{2}} + c \\ \frac{y}{x^2} &= \frac{3}{x^{2}} + c \\ y &= 3+cx^2 \end{align*}
d
x
d
u
u
x
2
y
y
=
−
x
3
6
=
x
2
3
+
c
=
x
2
3
+
c
=
3
+
c
x
2
When
x
=
1
,
y
=
2
{x=1, y=2}
x
=
1
,
y
=
2
2
=
3
(
1
)
+
c
(
1
)
2
c
=
−
1
\begin{gather*} 2 = 3(1)+c(1)^2 \\ c = -1 \end{gather*}
2
=
3
(
1
)
+
c
(
1
)
2
c
=
−
1
y
=
3
−
x
2
■
y = 3-x^2 \; \blacksquare
y
=
3
−
x
2
■
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