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AP Calculus BC
Unit 7: Differential Equations
7.5 Approximating Solutions Using Euler’s Method
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A differential equation relates a function to its
derivatives
An initial condition specifies the value of the
function
at a particular point.
What is the general form of a differential equation?
\frac{dy}{dx} = f(x, y)</latex>
Give an example of a differential equation.
d
y
d
x
=
\frac{dy}{dx} =
d
x
d
y
=
3
x
2
+
3x^{2} +
3
x
2
+
y
y
y
Give an example of an initial condition.
y
(
0
)
=
y(0) =
y
(
0
)
=
1
1
1
The step size in
Euler's Method
is denoted byh</latex>.
In Euler's Method,
y
n
y_{n}
y
n
is the approximated value of
y
y
y
at the current step
Match the variable with its meaning in Euler's Method:
y
n
+
1
y_{n + 1}
y
n
+
1
↔️ Approximated value of
y
y
y
at the next step
y
n
y_{n}
y
n
↔️ Approximated value of
y
y
y
at the current step
h
h
h
↔️ Step size
f
(
x
n
,
y
n
)
f(x_{n}, y_{n})
f
(
x
n
,
y
n
)
↔️ Value of the differential equation at the current coordinates
What is the formula for Euler's Method?
y_{n + 1} = y_{n} + h \cdot f(x_{n}, y_{n})</latex>
In Euler's Method, a smaller step size
h
h
h
generally leads to a more accurate approximation.
Given
d
y
d
x
=
\frac{dy}{dx} =
d
x
d
y
=
x
+
x +
x
+
y
y
y
,
y
(
0
)
=
y(0) =
y
(
0
)
=
1
1
1
, and
h
=
h =
h
=
0.1
0.1
0.1
, the value of
y
1
y_{1}
y
1
is 1.1
Steps to apply Euler's Method
1️⃣ Identify differential equation, initial condition, step size, and target x</latex>
2️⃣ Iterate using the formula
y
n
+
1
=
y_{n + 1} =
y
n
+
1
=
y
n
+
y_{n} +
y
n
+
h
⋅
f
(
x
n
,
y
n
)
h \cdot f(x_{n}, y_{n})
h
⋅
f
(
x
n
,
y
n
)
3️⃣ Increment
n
n
n
and update
x
n
x_{n}
x
n
4️⃣ Present results in a table
Approximate
y
(
0.2
)
y(0.2)
y
(
0.2
)
for
d
y
d
x
=
\frac{dy}{dx} =
d
x
d
y
=
x
+
x +
x
+
y
y
y
with
y
(
0
)
=
y(0) =
y
(
0
)
=
1
1
1
and
h
=
h =
h
=
0.1
0.1
0.1
.
1.22
Steps of Euler's Method algorithm
1️⃣ Initialize step size and target x
2️⃣ Iterate using the formula
3️⃣ Update x and y values
4️⃣ Present results in a table
What is the formula used in Euler's Method for each iteration?
y_{n + 1} = y_{n} + h \cdot f(x_{n}, y_{n})</latex>
In each iteration of Euler's Method, the value of
x
n
x_{n}
x
n
is updated using the formula:
x
n
=
x_{n} =
x
n
=
x
0
+
x_{0} +
x
0
+
n
⋅
n \cdot
n
⋅
h
What happens to computational effort when the step size in Euler's Method decreases?
It increases
For
d
y
d
x
=
\frac{dy}{dx} =
d
x
d
y
=
x
+
x +
x
+
y
y
y
with y(0) = 1</latex>, using
h
=
h =
h
=
0.01
0.01
0.01
provides a more accurate result than
h
=
h =
h
=
0.1
0.1
0.1
because the step size is smaller
Match the step size with its corresponding accuracy in Euler's Method:
Large step size ↔️ Low accuracy
Small step size ↔️ High accuracy
Steps to approximate y(0.2)</latex> using Euler's Method for
d
y
d
x
=
\frac{dy}{dx} =
d
x
d
y
=
x
+
x +
x
+
y
y
y
with
y
(
0
)
=
y(0) =
y
(
0
)
=
1
1
1
and
h
=
h =
h
=
0.1
0.1
0.1
1️⃣ Calculate
f
(
0
,
1
)
=
f(0, 1) =
f
(
0
,
1
)
=
1
1
1
2️⃣ Calculate
y
(
0.1
)
=
y(0.1) =
y
(
0.1
)
=
1
+
1 +
1
+
0.1
⋅
1
=
0.1 \cdot 1 =
0.1
⋅
1
=
1.1
1.1
1.1
3️⃣ Calculate
f
(
0.1
,
1.1
)
=
f(0.1, 1.1) =
f
(
0.1
,
1.1
)
=
1.2
1.2
1.2
4️⃣ Calculate
y
(
0.2
)
=
y(0.2) =
y
(
0.2
)
=
1.1
+
1.1 +
1.1
+
0.1
⋅
1.2
=
0.1 \cdot 1.2 =
0.1
⋅
1.2
=
1.22
1.22
1.22
What is the approximate value of
y
(
0.2
)
y(0.2)
y
(
0.2
)
using Euler's Method for
d
y
d
x
=
\frac{dy}{dx} =
d
x
d
y
=
x
+
x +
x
+
y
y
y
with
y
(
0
)
=
y(0) =
y
(
0
)
=
1
1
1
and
h
=
h =
h
=
0.1
0.1
0.1
?
1.22