Second order differentiel equations

Gamestops

New member
Joined
Dec 7, 2018
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2
Hey everybody

So I'm fairly new to differential equations and this is my first time dealing with one of second order. Therefore if the problem seems easy that's why.

I am supposed to rewrite or reduce the following equation to a system of 1. order differential equations

y''-0.1*y+2=0

and then find a numerical solution where the following is true
y(0)=10 and y'(0)=-1

I'm not necessarily looking for a solution, but rather if someone could forward me to some kind guide or lesson about this.
 
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Subhotosh Khan

Super Moderator
Staff member
Joined
Jun 18, 2007
Messages
18,241
Hey everybody

So I'm fairly new to differentiel equations and this is my first time dealing with one of second order. Therefore if the problem seems easy thats why.

I am supposed rewrite or reduce the following equation to a system of 1. order differentiel equations

y''*0,1*y+2=0 ................... What does that mean?

and then find a numerical solution where the following is true
y(0)=10 and y'(0)-1

i'm not necessarily looking for a solution, but rather if someone could forward me to some kind guide or lesson about this.
.
 

Gamestops

New member
Joined
Dec 7, 2018
Messages
2
Oh sorry forgot to convert it to american notation

it should say y''-0.1*y+2=0
 

Theodor

New member
Joined
Dec 7, 2018
Messages
3
Hey everybody

So I'm fairly new to differential equations and this is my first time dealing with one of second order. Therefore if the problem seems easy that's why.

I am supposed to rewrite or reduce the following equation to a system of 1. order differential equations

y''-0.1*y+2=0

and then find a numerical solution where the following is true
y(0)=10 and y'(0)=-1

I'm not necessarily looking for a solution, but rather if someone could forward me to some kind guide or lesson about this.
Z^2-0.1=0
y=C1exp(0.1x)+C2exp(-0.1x)
y(0)=C1+C2=10 (1)
y'(0)=0.1*C1-0.1*C2=-1 or y'(0)=C1-C2=-10 (2)
(1) and (2) => C1=0 C2=10
y= 10*exp(-0.1x)
 
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