MSU
Main Campus
Marawi City
ENGINEERING MATHEMATICS
COURSE CODE: ENS 181
INSTRUCTOR: Edilvin Walter P. Maghanoy
Order and Classification of Differential Equations
Differential equations are classified based on their order, which is determined by the highest
derivative present.
First-Order ODE:
𝑑𝑦
=𝑥+1
𝑑𝑥
Second-order ODE:
𝑑2 𝑦
𝑑𝑦
+
3
+ 2𝑦 = 0
𝑑𝑥 2
𝑑𝑥
We also classify differential equations as ordinary differential equations (ODEs) or partial differential equations (PDEs)
ODEs involve functions of a single variable
PDEs involve functions of multiple variables:
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𝑑2 𝑦
𝑑𝑦
+
3
+ 2𝑦 = 0
𝑑𝑥 2
𝑑𝑥
𝛿𝑇
𝛿2 𝑇
=𝑘 2
𝛿𝑡
𝛿𝑥
First-order ODE’s
In the real world, many processes can be described using different forms of first-order
ODEs. Some of these equations can be solved directly using algebraic manipulation,
while others require more advanced techniques.
1. Separable ODEs
2. Exact ODEs
3. Linear ODEs
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Separable ODE’s
Many differential equations in engineering applications can be written in a form that
allows us to separate the variables., meaning that all terms involving y appear on one
side and all terms involving x appear on the other
𝑔 𝑦
𝑑𝑦
=𝑓 𝑥
𝑑𝑥
𝑔 𝑦 𝑑𝑦 = 𝑓 𝑥 𝑑𝑥
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Recall: Population Growth
Solving Separable ODEs
1.
2.
3.
4.
Rewrite the equation to separate the variables.
Integrate both sides.
Introduce the constant of integration C.
Solve for y if possible
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Example 1
Solve the ODE 𝑦 ′ = 1 + 𝑦 2
Step 1: Separate the variables
𝑑𝑦
= 𝑑𝑥
2
1+𝑦
Step 3: Introduce constant C
𝑡𝑎𝑛−1 𝑦 = 𝑥 + C
Step 2: Integrate both sides
Step 4: Solve for y
𝑑𝑦
න
= න 𝑑𝑥
2
1+𝑦
𝑦 = tan(𝑥 + 𝐶)
𝑡𝑎𝑛−1 𝑦 = 𝑥
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Example 2
In 1991, scientists discovered the Iceman (Oetzi), a prehistoric mummy, in the Oetztal
Alps. Analysis shows that the ratio of carbon-14 to carbon-12 in his remains was 52.5%
of that in a living organism. Given the half-life of carbon-14 is 5715 years, estimate the
time since Oetzi’s death.
Understand the problem:
1. The ratio of carbon-14 to carbon-12 in living organisms is constant.
2. When an organism dies, it stops absorbing carbon-14.
3. The remaining carbon-14 decays exponentially, allowing scientist to estimate when
the organism died.
4. The half-life of carbon-14 is H = 5715 years.
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Example 2
Formulate the DE:
𝑑𝑦
𝑑𝑡
= 𝑘𝑦
Step 1: Separate the variables
𝑑𝑦
= 𝑘𝑑𝑡
𝑦
Step 2: Integrate both sides
𝑑𝑦
න
= න 𝑘𝑑𝑡
𝑦
ln |𝑦| = 𝑘𝑡 + 𝐶
Step 3: Solve for y
𝑦 = 𝑦0 𝑒 𝑘𝑡
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Where:
− 𝑦 𝑖𝑠 𝑡ℎ𝑒 𝑟𝑎𝑡𝑖𝑜 𝑜𝑓 14𝐶 to 12C.
- 𝑘 𝑖𝑠 𝑡ℎ𝑒 𝑑𝑒𝑐𝑎𝑦 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
- t is the time elapsed
Step 4: Determine k using half-life
0.5𝑦0 = 𝑦0 𝑒 𝑘𝑡
𝑒 𝑘𝑡 = 0.5
ln 0.5 −0.693
𝑘 = 𝐻 = 5715 = −0.0001213
Step 5: Calculate time t for Oetzi’s Death
Given that the carbon-14 ratio is 52.5%
𝑦 = 0.525𝑦0
𝑦 = 𝑦0 𝑒 𝑘𝑡
0.525𝑦0 = 𝑦0 𝑒 −0.0001213𝑡
ln 0.525 = −0.0001213𝑡
𝑡 = 5312 𝑦𝑒𝑎𝑟𝑠
Oetzi died approximately
5300 years ago.
Example 3
A 1000-gallon tank initially contains 100 lbs of salt. Brine flows in at 10 gal/min with 5
lbs of salt per gallon, while an equal amount exits. Find the salt content in the tank
over time.
Step 1: Setting up a model
Let 𝑦 𝑡 denote the amount of salt in the tank at time t
𝒅𝒚
= 𝑺𝒂𝒍𝒕 𝒊𝒏𝒇𝒍𝒐𝒘 𝑹𝒂𝒕𝒆 − 𝑺𝒂𝒍𝒕 𝑶𝒖𝒕𝒇𝒍𝒐𝒘 𝑹𝒂𝒕𝒆
𝒅𝒕
Inflow:
𝑙𝑏𝑠
Brine enters at 10 𝑔𝑎𝑙/𝑚𝑖𝑛, with 5 𝑔𝑎𝑙 salt concentration
Outflow:
Tank volume is 1000 𝑔𝑎𝑙, so salt concentration at time t is:
𝑦
1000
𝑙𝑏𝑠/𝑔𝑎𝑙
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𝑔𝑎𝑙
Brine leaves at 10
, so salt outflow is
𝑚𝑖𝑛
𝑦
10 𝑥
= 0.01𝑦 𝑙𝑏𝑠/𝑚𝑖𝑛
1000
So, the differential equation for the
amount of salt y(t) is:
𝑑𝑦
= 50 − 0.01𝑦
𝑑𝑡
Example 3
A 1000-gallon tank initially contains 100 lbs of salt. Brine flows in at 10 gal/min with 5
lbs of salt per galoon, while an equal amount exits. Find the salt content in the tank
over time.
So, the differential equation for the
amount of salt y(t) is :
𝑑𝑦
= 50 − 0.01𝑦
𝑑𝑡
𝑑𝑦
= −0.01𝑑𝑡
𝑦 − 5000
Integrating both sides:
𝑑𝑦
න
= න −0.01𝑑𝑡
𝑦 − 5000
ln y − 5000 = −0.01t + C
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Exponentiate both sides:
Apply Initial Conditions:
𝑒 ln |𝑦−5000| = 𝑒 −0.01𝑡+𝐶
𝑦 − 5000 = 𝑒 𝐶 𝑒 −0.01𝑡
y − 5000 = 𝐶𝑒 −0.01𝑡
𝒚 = 𝟓𝟎𝟎𝟎 + 𝑪𝒆−𝟎.𝟎𝟏𝒕
y = 5000 + 𝐶𝑒 −0.01𝑡
100 = 5000 + 𝐶𝑒 0
100 = 5000 + 𝐶
C = −4900
Apply Initial Conditions:
y t = 5000 − 4900𝑒 −0.01𝑡
𝑦 0 = 100
Initially, the tank contains
100 lbs of salt
Example 3
A 1000-gallon tank initially contains 100 lbs of salt. Brine flows in at 10 gal/min with 5
lbs of salt per galoon, while an equal amount exits. Find the salt content in the tank
over time.
y t = 5000 − 4900𝑒 −0.01𝑡
Interpretation:
At t -> ∞, y -> 5000
This means the salt content stabilizes at 5000 lbs over time
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Reduction to Separable Form
Solve 2𝑥𝑦𝑦 ′ = 𝑦 2 − 𝑥 2
Can be expressed in the standard form
2
2
𝑦
−
𝑥
𝑦′ =
2𝑥𝑦
In many cases, non-separable differential equations can be transformed into
separable form by introducing a new variable.
𝑢=
𝑦
𝑥
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𝑦 ′ = 𝑢′ 𝑥 + 𝑢
𝑦 = 𝑢𝑥, substitute this to the equation
Reduction to Separable Form
Solve 2𝑥𝑦𝑦 ′ = 𝑦 2 − 𝑥 2
Can be expressed in the standard form
𝑦 2 −𝑥 2
′
𝑦=
2𝑥𝑦
=
𝑦
2𝑥
−
𝑥
2𝑦
2
𝑢
1
𝑢 −1
′
𝑢 𝑥+𝑢 =
−
=
2 2𝑢
2𝑢
𝑦 ′ = 𝑢′𝑥 + 𝑢
𝑦 = 𝑢𝑥
Separating variables
2𝑢𝑑𝑢
𝑑𝑥
1
2
by Integration ln 1 + 𝑢 = − ln 𝑥 + 𝐶 = ln + 𝐶
2 = −
1+𝑢
𝑒
ln 1+𝑢2
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𝑥
=𝑒
𝑥
ln
1
+𝐶
𝑥
𝑥 2 + 𝑦 2 = 𝐶𝑥