Identify differential equations

[Incubator]

I have some equation which i need to learn for my exam, but i'm having trouble identifying them, and i can't search for them on the internet before i know what kind of equations they are, it would mean a lot to me if someone could identify them.

1. y' = sin(x + y + 1)

2. y'₁ = 3y₁ - 2y₂
y'₂= 2y₁ - 2y₂

3. y⁽³⁾ - y⁽¹⁾ = x + e^​2x + cosx(3x)

 

[HallsofIvy]

I have some equation which i need to learn for my exam, but i'm having trouble identifying them, and i can't search for them on the internet before i know what kind of equations they are, it would mean a lot to me if someone could identify them.

1. y' = sin(x + y + 1)

This is a first order differential equations since it only involves the first derivative of y. It looks to me like it would be very difficult to solve!

2. y'₁ = 3y₁ - 2y₂
y'₂= 2y₁ - 2y₂

This is a "system of linear differential equations". Unlike the first, this is fairly easy to solve for y₁ and y₂

3. y⁽³⁾ - y⁽¹⁾ = x + e^​2x + cosx(3x)

This is a "third order linear non-homogeneous differential equation with constant coefficients" (the numbers in parentheses indicate derivatives). It's "characteristic equation" is [itex]\lambda^3- \lambda= 0[/itex] so has 0, 1, and -1 as characteristic roots. Once you have found the general solution to the associated homogeneous equation (\(\displaystyle y^{(3)}- y^{(1)}= 0\)), you can look for values of A, B, C and D so that \(\displaystyle Ax+ B+ Ce^{2x}+ Dcos(3x)\) satisfies the entire equation and add that.

I'm very concerned that you need to "learn" these for an exam and don't know what kind of equations they are. That's peculiar.

 

[Incubator]

Thanks, i know it sounds bad, but i don't really like differential equations, because in my case i will never really use it, so i didn't pay attention. But these are equations given in the exams, and my only source to learn them would be the internet :\

 

[Deleted member 4993]

This is a first order differential equations since it only involves the first derivative of y. It looks to me like it would be very difficult to solve!

This is a "system of linear differential equations". Unlike the first, this is fairly easy to solve for y₁ and y₂


This is a "third order linear non-homogeneous differential equation with constant coefficients" (the numbers in parentheses indicate derivatives). It's "characteristic equation" is [itex]\lambda^3- \lambda= 0[/itex] so has 0, 1, and -1 as characteristic roots. Once you have found the general solution to the associated homogeneous equation (\(\displaystyle y^{(3)}- y^{(1)}= 0\)), you can look for values of A, B, C and D so that \(\displaystyle Ax+ B+ Ce^{2x}+ Dcos(3x)\) satisfies the entire equation and add that.

I'm very concerned that you need to "learn" these for an exam and don't know what kind of equations they are. That's peculiar.

The only thing I would add:

The first equation is a non-linear ODE, where as the second and the third one are linear ODE.

 

[Incubator]

So, i tried to solve the second problem...

i will call y1 => x and y2 => y to make it easier to write

so we will have
x' = 3x - 2y
y' = 2x - 2y

2y = 3x - x'
y = (3x - x')/2
y' = (3x' - x'')/2

(3x' - x'')/2 = 2x - 2(3x - x')/2 | (*2)
3x' - x'' = 4x - 2(3x - x')
3x' - x'' = 4x - 6x - 2x'
-x'' + 5x' + 2x = 0

I'm not sure what to do from here if this is alright

 

[Deleted member 4993]

So, i tried to solve the second problem...

i will call y1 => x and y2 => y to make it easier to write

so we will have
x' = 3x - 2y
y' = 2x - 2y

2y = 3x - x'
y = (3x - x')/2
y' = (3x' - x'')/2

(3x' - x'')/2 = 2x - 2(3x - x')/2 | (*2)
3x' - x'' = 4x - 2(3x - x')
3x' - x'' = 4x - 6x - 2x'
-x'' + 5x' + 2x = 0

I'm not sure what to do from here if this is alright

This is a standard homogeneous second order linear differential equation.

Find the characteristic equation for this ODE.

If you don't know what that means - do a google search and tell us what you found.

 

[Lost souls]

This is a first order differential equations since it only involves the first derivative of y. It looks to me like it would be very difficult to solve!

cant it be solved like this

(dy/dx)= sin(x+y+1)
let (x+y+1)=t;

(dt/dx)-1= sin t
(dt/dx)=sin t +1
int(dx)= int(dt/(sint +1))

???

 

[Deleted member 4993]

cant it be solved like this

(dy/dx)= sin(x+y+1)

No!

let (x+y+1)=t; Now t is function of both x and y.

(dt/dx)-1= sin t
(dt/dx)=sin t +1
int(dx)= int(dt/(sint +1))

???

y' = sin(x+y+1)

y" = cos(x+y+1) * (1 + y')

y'" = y" * cos(x+y+1) - sin(x+y+1) * (1+y') = y" * y"/(1+y') - y' (1+y')

and continue.....

(By the way, Wolfram-alpha timed out...)