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Solving a first order linear IVP

LLand314

Member
Mar 26, 2013
77
Re: Logan's question at Yahoo! Answers involving an IVP with a linear 1st order ODE

How would I solve a DE like
t(dy/dt) + 7y = t^3

dy/dt +7y/t = t^2
Then do I need to find the integrating factor?
 
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MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: Logan's question at Yahoo! Answers involving an IVP with a linear 1st order ODE

Yes, your next step is to compute the integrating factor. Can you state what it needs to be, in integral form?
 
Last edited:

LLand314

Member
Mar 26, 2013
77
Hey no worries, I was just showing you a better way to write the integrating factor, just for clarity. :D

Yes, your next step is to compute the integrating factor. Can you state what it needs to be, in integral form?

Note: For any additional questions, I would encourage you to begin a new topic in the Differential Equations forum. It is preferable to begin a new topic for a new question so that topics do not become potentially cluttered and hard to follow.

I am happy you have registered and are participating here! (Yes)
I'm not sure what the integrating factor is but I feel like it has to do with ln since its 7y/t
Maybe e^7ln(y/t) ? (Just guessing)


Thanks
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: Logan's question at Yahoo! Answers involving an IVP with a linear 1st order ODE

You are close, but in this case we have:

\(\displaystyle P(t)=\frac{7}{t}\) and so our integrating factor is:

\(\displaystyle e^{7\int\frac{dt}{t}}\)

Can you evaluate this and simplify?
 

LLand314

Member
Mar 26, 2013
77
Re: Logan's question at Yahoo! Answers involving an IVP with a linear 1st order ODE

You are close, but in this case we have:

\(\displaystyle P(t)=\frac{7}{t}\) and so our integrating factor is:

\(\displaystyle e^{7\int\frac{dt}{t}}\)

Can you evaluate this and simplify?
Would that be |t^7|
So the integrating factor is t^7?
Or did I do that completely wrong
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: Logan's question at Yahoo! Answers involving an IVP with a linear 1st order ODE

You are absolutely correct! (Yes)

So, now your next step is to multiply the ODE by this integrating factor, and observe the left side is the product of the derivative of the integrating factor and the dependent variable $y$. Can you show this?
 

LLand314

Member
Mar 26, 2013
77
Re: Logan's question at Yahoo! Answers involving an IVP with a linear 1st order ODE

You are absolutely correct! (Yes)

So, now your next step is to multiply the ODE by this integrating factor, and observe the left side is the product of the derivative of the integrating factor and the dependent variable $y$. Can you show this?

t^7(dy/dt) + (7y/t)t^7 = t^2 t^7

(d/dx) t^7 = t^14 ?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: Logan's question at Yahoo! Answers involving an IVP with a linear 1st order ODE

On the left you may simplify using \(\displaystyle \frac{t^7}{t}=t^6\) and on the right, recall the rule:

\(\displaystyle a^b\cdot a^c=a^{b+c}\)

so that the right side becomes \(\displaystyle t^{2+7}=t^9\)

and so you now have:

\(\displaystyle t^7\frac{dy}{dt}+7t^6y=t^9\)

Now, how can we rewrite the left side?
 

LLand314

Member
Mar 26, 2013
77
Re: Logan's question at Yahoo! Answers involving an IVP with a linear 1st order ODE

On the left you may simplify using \(\displaystyle \frac{t^7}{t}=t^6\) and on the right, recall the rule:

\(\displaystyle a^b\cdot a^c=a^{b+c}\)

so that the right side becomes \(\displaystyle t^{2+7}=t^9\)

and so you now have:

\(\displaystyle t^7\frac{dy}{dt}+7t^6y=t^9\)

Now, how can we rewrite the left side?

Do we write it as a differentiation of a product like my first problem?

(d/dt) t^7y = t^9
Then integrate with respect to t?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: Logan's question at Yahoo! Answers involving an IVP with a linear 1st order ODE

Yes, although, if I may suggest, this notation would be preferable:

d/dt(t^7·y) = t^9.

And yes, now you want to integrate with respect to $t$:

\(\displaystyle \int\,d\left(t^7y \right)=\int t^9\,dt\)
 

LLand314

Member
Mar 26, 2013
77
Re: Logan's question at Yahoo! Answers involving an IVP with a linear 1st order ODE

Yes, although, if I may suggest, this notation would be preferable:

d/dt(t^7·y) = t^9.

And yes, now you want to integrate with respect to $t$:

\(\displaystyle \int\,d\left(t^7y \right)=\int t^9\,dt\)
integral d(t^7·y) = integral t^9 dt
t^7·y = t^10/10
y= t^3/10 + Ct^-7
y(2)=1
When I plug in the values I get a crazy number is my y not correct?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: Logan's question at Yahoo! Answers involving an IVP with a linear 1st order ODE

Your general solution is correct. What value do you obtain for the parameter $C$?
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,191
Re: Logan's question at Yahoo! Answers involving an IVP with a linear 1st order ODE

How would I solve a DE like
t(dy/dt) + 7y = t^3

dy/dt +7y/t = t^2
Then do I need to find the integrating factor?
Another option is to recognize that this equation is Cauchy-Euler.
 

LLand314

Member
Mar 26, 2013
77
Re: Logan's question at Yahoo! Answers involving an IVP with a linear 1st order ODE

Your general solution is correct. What value do you obtain for the parameter $C$?

I think I figured C out correctly.
y= t^3/10 - (128/5)t^-7

- - - Updated - - -

Thanks for sticking with me and helping me through to the end.

Thanks again.
 
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MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
I get \(\displaystyle C=\frac{128}{5}\) and so the solution satisfying the IVP is:

\(\displaystyle y(t)=\frac{t^3}{10}+\frac{128}{5t^7}\)
 

LLand314

Member
Mar 26, 2013
77
I get \(\displaystyle C=\frac{128}{5}\) and so the solution satisfying the IVP is:

\(\displaystyle y(t)=\frac{t^3}{10}+\frac{128}{5t^7}\)

Ah yes my mistake I meant + but yes I got c=128/5
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: Logan's question at Yahoo! Answers involving an IVP with a linear 1st order ODE

Another option is to recognize that this equation is Cauchy-Euler.
That's a nice observation! Since when I was a student we were not shown this method until studying higher order equations, I will demonstrate it for the OP.

We are originally given:

\(\displaystyle t\frac{dy}{dt}+7y=t^3\)

If we make the substitution:

\(\displaystyle t=e^x\) it then follows from the chain rule that:

\(\displaystyle \frac{dy}{dx}=\frac{dy}{dt}\frac{dt}{dx}=\frac{dy}{dt}e^x=t\frac{dy}{dt}\)

and our ODE becomes:

\(\displaystyle \frac{dy}{dx}+7y=e^{3x}\)

We find our integrating factor is \(\displaystyle \mu(x)=e^{7x}\):

\(\displaystyle e^{7x}\frac{dy}{dx}+7e^{7x}y=e^{10x}\)

\(\displaystyle \frac{d}{dx}\left(e^{7x}y \right)=e^{10x}\)

\(\displaystyle e^{7x}y=\frac{1}{10}e^{10x}+C\)

\(\displaystyle y(x)=\frac{e^{3x}}{10}+\frac{C}{e^{7x}}\)

Now, back-substituting for $x$ we obtain:

\(\displaystyle y(t)=\frac{t^3}{10}+\frac{C}{t^7}\)
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,191
Another method is simply to assume a solution of the form $y=t^{n}$. Then plug in $\dot{y}=n t^{n-1}$, so the associated homogeneous DE becomes
$$t(n t^{n-1})+7t^{n}=0,$$
or
$$t^{n}(n+7)=0.$$
For $t\not=0$, you must have $n=-7$, and then $y_{h}=A t^{-7}$. For a particular solution, assume $y_{p}=B t^{3}$, and you'll get by plugging in that $y_{p}=t^{3}/10$, and you get the solution you got by the change of variables.