Could someone just give me a quick step by step guide on how to work the Initial Value Problem on first order differential equations. I'm kind of lost with it right now :/
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Initial Value Problem
Things to Note:
- you are given the initial condition
- you are given a differential equation
- the solution you come up with should satisfy the given conditions
simple problem (y = 2x + 6 , this will be our solution that will be determined from the given IVP)
`````````````````````
solve the differential equation y '(x) = 2 given y(0) = 6
procedure:
kind of separable
dy/dx = 2
dy = 2 dx
integrate
∫dy = ∫2 dx
y = 2x + C
C is just an arbitrary constant
Use the conditions given, that is y(0) = 6 means when x = 0, y = 6; plug in these values to determine C
y = 2x + C
6 = 2(0) + C
C = 6
solution:
y = 2x + C
y = 2x + 6
```````````````
slightly harder problem;( y = 2e^(x) + 5e^(3x) , will be the solution that will be determined from the given IVP)
y'' - 4y' + 3y = 0, y(0) = 7, y '(0) = 17
characteristic equation(we calculate the roots)
r² - 4r + 3 = 0
r² - 3r - r + 3 = 0
r(r - 3) - 1(r - 3) = 0
r = 1 and 3
r₁ = 1 and r₂ = 3
general solution
y = C₁e^(r₁x) + C₂e^(r₂x)
y = C₁e^(x) + C₂e^(3x) this would be the general solution to the differential equation y'' - 4y' + 3y = 0 .
The other conditions are included so that we can determine the solution to the IVP.
y(0) = 7 means x = 0, y = 7
y = C₁e^(x) + C₂e^(3x)
7 = C₁e^(0) + C₂e^(0)
7 = C₁ + C₂
~~~~~~~~
y '(0) = 17 means the first derivative of the general solution y = C₁e^(x) + C₂e^(3x) is equal to 17 when x is 0
y = C₁e^(x) + C₂e^(3x)
y ' = C₁e^(x) + 3C₂e^(3x)
17 = C₁e^(0) + 3C₂e^(0)
17 = C₁ + 3C₂
~~~~~~~~~~~
solve for C's simultaneously
17 = C₁ + 3C₂
7 = C₁ + C₂
I use elimination method to eliminate C₁
17 - 7 = C₁ - C₁ + 3C₂ - C₂
10 = 2C₂
C₂ = 5
````````
7 = C₁ + C₂
7 = C₁ + 5
C₁ = 2
````````
Substitute these values in the general solution to solve the IVP, y'' - 4y' + 3y = 0, y(0) = 7, y '(0) = 17.
y = C₁e^(x) + C₂e^(3x)
y = 2e^(x) + 5e^(3x) is the solution to the IVP y'' - 4y' + 3y = 0, y(0) = 7, y '(0) = 17.
``````````````````````````
Things to Note:
- you are given the initial condition
- you are given a differential equation
- the solution you come up with should satisfy the given conditions
simple problem (y = 2x + 6 , this will be our solution that will be determined from the given IVP)
`````````````````````
solve the differential equation y '(x) = 2 given y(0) = 6
procedure:
kind of separable
dy/dx = 2
dy = 2 dx
integrate
∫dy = ∫2 dx
y = 2x + C
C is just an arbitrary constant
Use the conditions given, that is y(0) = 6 means when x = 0, y = 6; plug in these values to determine C
y = 2x + C
6 = 2(0) + C
C = 6
solution:
y = 2x + C
y = 2x + 6
```````````````
slightly harder problem;( y = 2e^(x) + 5e^(3x) , will be the solution that will be determined from the given IVP)
y'' - 4y' + 3y = 0, y(0) = 7, y '(0) = 17
characteristic equation(we calculate the roots)
r² - 4r + 3 = 0
r² - 3r - r + 3 = 0
r(r - 3) - 1(r - 3) = 0
r = 1 and 3
r₁ = 1 and r₂ = 3
general solution
y = C₁e^(r₁x) + C₂e^(r₂x)
y = C₁e^(x) + C₂e^(3x) this would be the general solution to the differential equation y'' - 4y' + 3y = 0 .
The other conditions are included so that we can determine the solution to the IVP.
y(0) = 7 means x = 0, y = 7
y = C₁e^(x) + C₂e^(3x)
7 = C₁e^(0) + C₂e^(0)
7 = C₁ + C₂
~~~~~~~~
y '(0) = 17 means the first derivative of the general solution y = C₁e^(x) + C₂e^(3x) is equal to 17 when x is 0
y = C₁e^(x) + C₂e^(3x)
y ' = C₁e^(x) + 3C₂e^(3x)
17 = C₁e^(0) + 3C₂e^(0)
17 = C₁ + 3C₂
~~~~~~~~~~~
solve for C's simultaneously
17 = C₁ + 3C₂
7 = C₁ + C₂
I use elimination method to eliminate C₁
17 - 7 = C₁ - C₁ + 3C₂ - C₂
10 = 2C₂
C₂ = 5
````````
7 = C₁ + C₂
7 = C₁ + 5
C₁ = 2
````````
Substitute these values in the general solution to solve the IVP, y'' - 4y' + 3y = 0, y(0) = 7, y '(0) = 17.
y = C₁e^(x) + C₂e^(3x)
y = 2e^(x) + 5e^(3x) is the solution to the IVP y'' - 4y' + 3y = 0, y(0) = 7, y '(0) = 17.
``````````````````````````