MATEMÀTIQUES PER L’ENGINYERIA
1r PARCIAL
First-order differential equations
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.
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.
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)𝑎𝑝𝑝𝑒𝑎𝑟𝑠𝑜𝑛𝑦ℎ (𝑡)𝑤𝑒ℎ𝑎𝑣𝑒𝑎𝑝𝑟𝑜𝑏𝑙𝑒𝑚! 𝐼𝑡𝑤𝑖𝑙𝑙𝑏𝑒𝑧𝑒𝑟𝑜, 𝑠𝑜𝑤𝑒𝑤𝑖𝑙𝑙ℎ𝑎𝑣𝑒𝑡𝑜𝑎𝑑𝑑𝑡𝑡𝑜𝑡ℎ𝑒𝑔𝑢𝑒𝑠𝑠
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, 𝑤ℎ𝑒𝑟𝑒𝑌(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
0
