How Do You Solve an Initial Value Problem with Integration Steps?

[newtomath]

I need help with an initial value problem,

ty' + (t+1)y= t; y (LN 2)= 1

I divided t and have u(t) as exp Integral of t+1/1 => e^t +t

Multiplied this to the original equation to get

(e^t +t)y' + ((t+ 1)/t) *y *(e^t +t) = (e^t +t)

How can I integrate the above? Are my steps so far correct?

Thanks

 

[JJacquelin]

I divided t and have u(t) as exp Integral of t+1/1 => e^t +t

( e^t + t ) isn't correct !

 

[shelovesmath]

I need help with an initial value problem,

ty' + (t+1)y= t; y (LN 2)= 1

I divided t and have u(t) as exp Integral of t+1/1 => e^t +t

Multiplied this to the original equation to get

(e^t +t)y' + ((t+ 1)/t) *y *(e^t +t) = (e^t +t)

How can I integrate the above? Are my steps so far correct?

Thanks

So, you are trying to solve using linear method?

 

[HallsofIvy]

I need help with an initial value problem,

ty' + (t+1)y= t; y (LN 2)= 1

I divided t and have u(t) as exp Integral of t+1/1 => e^t +t

You mean that, to find an integrating factor, you integrated (t+1)/t which is the same as 1+ 1/t. From that you get ln(u)= t+ ln(t), $u= e^{t+ ln(t)}= e^t*e^{ln t}= te^t$, NOT $e^t+ t$

Multiplied this to the original equation to get

(e^t +t)y' + ((t+ 1)/t) *y *(e^t +t) = (e^t +t)

The whole point of the integrating factor is that the left side should be equal to
((e^t+ t)y)'- and it isn't!

How can I integrate the above? Are my steps so far correct?

Thanks

 

[newtomath]

Thanks for all the feedback.

@halls- thanks, te^t did it for me, problem solved.