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MATEMÀTIQUES PER L’ENGINYERIA
1r PARCIAL
First-order differential equations
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Modeling via Differential Equations: The course is about how to predict the future. So,
phrases such as ”rate of change of ...” or ”velocity” (derivative of position) lead to a
differential equations.
Classification by type:
1. Ordinary Differential equations (ODE): Only one independent variable
2. Partial Differential Equations (PDE): More independent variables
Classification by order: The order of a differential equation is the largest derivative
present in the differential equation.
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Autonomous: A differential equation not depending on x, independent variable, is
called autonomous.
𝑦 ′′ − 𝑦 + 𝑦 3 = 0
Linear: A differential equation is said to be linear if F can be written as a linear
combination of the derivatives of y. IT CAN’T HAVE POWERS!
𝑦 ′′ = 𝑦 + 𝑘sin⁡(𝑥)
Checking that a given function is a solution to a given equation:
𝑑𝑦 𝑦 2 − 1
=
𝑑𝑡 𝑡 2 + 2𝑡
𝑦1 (𝑡) = 1 + 𝑡; ⁡⁡𝑦2 (𝑡) = 1 + 2𝑡;⁡⁡𝑦3 (𝑡) = 1 Are solutions.
I.
Compute the left-hand side by differentiating 𝑦1 (𝑡)
𝑑𝑦1 𝑑(1 + 𝑡)
=
=1
𝑑𝑡
𝑑𝑡
Substituting 𝑦1 (𝑡) into the right-hand side:
𝑦 2 − 1 (𝑡 + 1)2 − 1
=
=1
𝑡 2 + 2𝑡
𝑡 2 + 2𝑡
III.
Both sides of the differential equation are identical, so is a solution.
First Order Differential Equations, Separables: Las variables dependientes e
independientes puedan ser agrupadas y, a partir de allí intentar una integración de
cada grupo por separado.
𝑑𝑦(𝑡)
𝑑𝑦
𝑑𝑦
= 𝑌(𝑦(𝑡))𝑇(𝑡) →
= 𝑇(𝑡)𝑑𝑡 ↔ ∫
= ∫ 𝑇(𝑡)𝑑𝑡
𝑑𝑡
𝑌(𝑦)
𝑌(𝑦)
II.
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First Order Differential Equations,Linear and Homogeneous: It can be written as a
separable one
𝑑𝑦
𝑑𝑦
= 𝑎(𝑡)𝑦 →
= 𝑎(𝑡)𝑑𝑡
𝑑𝑡
𝑦
Also general Solution: 𝑦(𝑡) = 𝐾𝑒 ∫ 𝑎(𝑡)𝑑𝑡
MATEMÀTIQUES PER L’ENGINYERIA

1r PARCIAL
First Order Differential Equations Lineal, NON Homogeneous:
𝑑𝑦
We have 𝑑𝑡 = 𝑎(𝑡)𝑦 + 𝑏(𝑡)
If the solution of the solution of the homogeneous part is not zero, then
𝑘𝑦ℎ (𝑡) + 𝑦𝑝 (𝑡) is the general solution.
DO GUESSINGS FOR Yp(t) if a(t) Is constant!
If 𝑦𝑝 (𝑡)⁡𝑎𝑝𝑝𝑒𝑎𝑟𝑠⁡𝑜𝑛⁡𝑦ℎ (𝑡)⁡𝑤𝑒⁡ℎ𝑎𝑣𝑒⁡𝑎⁡𝑝𝑟𝑜𝑏𝑙𝑒𝑚! 𝐼𝑡⁡𝑤𝑖𝑙𝑙⁡𝑏𝑒⁡𝑧𝑒𝑟𝑜⁡, 𝑠𝑜⁡𝑤𝑒⁡𝑤𝑖𝑙𝑙⁡ℎ𝑎𝑣𝑒⁡𝑡𝑜⁡𝑎𝑑𝑑⁡𝑡⁡𝑡𝑜⁡𝑡ℎ𝑒⁡𝑔𝑢𝑒𝑠𝑠⁡
If a(t) is not constant we have to apply the rules to obtain the general solution, we can’t
to the upper guess!
𝑑𝑦
= 𝑎(𝑡)𝑦 + 𝑏(𝑡)
𝑑𝑡
𝑑𝑦
− 𝑎(𝑡)𝑦 = 𝑏(𝑡)
𝑑𝑡
𝑑𝑦
𝑑𝑡
STEP 1
+ 𝑔(𝑡)𝑦 = 𝑏(𝑡)⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡𝑔(𝑡) = −𝑎(𝑡)
𝜇(𝑡) = 𝑒 ∫ 𝑔(𝑡)𝑑𝑡
STEP 2:
Define 𝛍(t)
𝑑𝑦
STEP 3:⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡𝜇(𝑡)(
Multiply
𝑑𝑡
both sides by 𝛍(t)
+ 𝑔(𝑡)𝑦) = 𝑏(𝑡)𝜇(𝑡)
𝑑
STEP
4: Integrate
⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡∫
((𝜇(𝑡) ∗ 𝑦(𝑡))
𝑑𝑡
both sides
= ∫ 𝑏(𝑡)𝜇(𝑡) 𝑑𝑡
⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡𝜇(𝑡)𝑦(𝑡) = ∫ 𝑏(𝑡)𝜇(t)dt
STEP 5: General
Solution
1
𝑦(𝑡) = 𝜇(𝑡) ∫ 𝜇(𝑡)𝑏(𝑡)𝑑𝑡
MATEMÀTIQUES PER L’ENGINYERIA

1r PARCIAL
Linear systems with Ktant coefficients:
We will have
𝑑𝑥
= 𝑎𝑥 + 𝑏𝑦
𝑑𝑡
𝑑𝑦
= 𝑐𝑥 + 𝑑𝑦
𝑑𝑡
Step 1: MATRIX A
𝑎
𝐴=(
𝑐
𝑏
)
𝑑
⁡𝑑𝑌
𝑥(𝑡)
⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡
= 𝐴𝑌⁡⁡⁡, 𝑤ℎ𝑒𝑟𝑒⁡𝑌(𝑡) = (
)
𝑦(𝑡)
𝑑𝑡
Step 2: Left hand side: derivem la solució que ens
donen, x(t)dt I y(t)dt I igualem a AY I comrovem si
són iguals both sides
Trobar la solució del Sistema given the initial value problem:
𝑥(𝑡)
Trobar K1 i k2 si 𝑌(𝑡) = (
) = 𝐾1 𝑦1 (𝑡) + 𝐾2 𝑦2 (𝑡)
𝑦(𝑡)
GENERAL SOLUTION