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Moved from a technical forum, so homework template missing

Hi again,

The previous problem was done using y′′(t)+2y′(t)+10y(t)=10 with with intial condition y(0⁻)=0.

In the following case, I'm using an initial condition and setting the right hand side equal to zero.

Find y(t) for the following differential equation with intial condition y(0⁻)=4.

y′′(t)+2y′(t)+10y(t)=0

Using the following formulas.

ℒy(t)=Y(s)

ℒ{y′(t)}=sY(s)-f(0⁻)=sY(s)-4

ℒ{y′′(t)}=ℒ{(sY(s)-4)′}=ℒ{(sY(s)′-4′)}=ℒ{(sY(s)′)}=s²Y(s)-4s

y′′(t)+2y′(t)+10y(t)=0

⇒{s²Y(s)-4s}+2{sY(s)-4}+10Y(s)=0

⇒s²Y(s)-4s+2sY(s)-8+10Y(s)=0

⇒s²Y(s)+2sY(s)+10Y(s)-4s-8=0

⇒s²Y(s)+2sY(s)+10Y(s)=4s+8

⇒Y(s){s²+2s+10}=4s+8

⇒Y(s) = (4s+8) / (s²+2s+10)

Now finding the inverse Laplace transform.

https://www.physicsforums.com/attac...261154/?hash=ab39a4f0474f3384a6a5c37cdc441e05

This method I found from this video.

You can see that I'm getting different functions, f(t), for the mentioned methods. Where am I going wrong? Which f(t) is correct? Thank you for the help.

The previous problem was done using y′′(t)+2y′(t)+10y(t)=10 with with intial condition y(0⁻)=0.

In the following case, I'm using an initial condition and setting the right hand side equal to zero.

Find y(t) for the following differential equation with intial condition y(0⁻)=4.

y′′(t)+2y′(t)+10y(t)=0

Using the following formulas.

ℒy(t)=Y(s)

ℒ{y′(t)}=sY(s)-f(0⁻)=sY(s)-4

ℒ{y′′(t)}=ℒ{(sY(s)-4)′}=ℒ{(sY(s)′-4′)}=ℒ{(sY(s)′)}=s²Y(s)-4s

y′′(t)+2y′(t)+10y(t)=0

⇒{s²Y(s)-4s}+2{sY(s)-4}+10Y(s)=0

⇒s²Y(s)-4s+2sY(s)-8+10Y(s)=0

⇒s²Y(s)+2sY(s)+10Y(s)-4s-8=0

⇒s²Y(s)+2sY(s)+10Y(s)=4s+8

⇒Y(s){s²+2s+10}=4s+8

⇒Y(s) = (4s+8) / (s²+2s+10)

Now finding the inverse Laplace transform.

**Method 1:**https://www.physicsforums.com/attac...261154/?hash=ab39a4f0474f3384a6a5c37cdc441e05

**
**

Method 2:Method 2:

This method I found from this video.

**
**

Question:Question:

You can see that I'm getting different functions, f(t), for the mentioned methods. Where am I going wrong? Which f(t) is correct? Thank you for the help.