OpenStudy (loser66):

find the solution of the given initial value problem $t^3y'+4t^2y =e^{-t}$y(-1) =0, t<0

OpenStudy (loser66):

$y = \frac{1}{t^4}\int t^3e^{-t}dt + C\frac{1}{t^4}$

OpenStudy (anonymous):

i'm assuming you solved that using integrating factors, right?

OpenStudy (loser66):

I got that, and use tabular method to take that integral, but I am wrong,

OpenStudy (loser66):

yes

OpenStudy (loser66):

would you please check my y?

OpenStudy (anonymous):

give me a sec

OpenStudy (anonymous):

i just had my final exam today for this class so it's pretty fresh on my mind haha

OpenStudy (loser66):

I saw you solve it before. today is my first day for this course. ha!! 2 chapters/ day . we are gonna fail......

OpenStudy (anonymous):

so your integral is a little wrong... it should be$\int\limits_{}^{}te^{-t}$ not $\int\limits_{}^{}t^{3}e^{-t}$

OpenStudy (loser66):

$e^{\int P(t)dt}= t^4$

OpenStudy (anonymous):

correct

OpenStudy (loser66):

and $e^{-\int P(t)dt}= \frac{1}{t^4}$

OpenStudy (loser66):

correct?

OpenStudy (anonymous):

yes, but what is the need for that?

OpenStudy (loser66):

because of the formula need them to plug in

OpenStudy (loser66):

$y = e^{-\int(P(t)dt}\int(g(t)e^{\int P(t)dt})dt + Ce^{-\int P(t)dt}$

OpenStudy (anonymous):

$e^{\int\limits\limits_{}^{}P(t)dt}y = \int\limits\limits_{}^{}e^{\int\limits\limits_{}^{}P(t)dt} *e^{-t}*t^{-3}dt$

OpenStudy (loser66):

I know this method, too. I watched videos on youtube. Ok, go ahead friend.

OpenStudy (anonymous):

well, you have $t^{4}y = \int\limits{}^{}e^{-t}tdt$

OpenStudy (anonymous):

integrate by parts, then divide your t^4 to solve for y

OpenStudy (loser66):

solve by your way: $t^3y' +4t^2y = e^{-t}$divide both sides by t^3 $y' +\frac{4}{t}y = \frac{e^{-t}}{t}$ integrating fractor is t^4 , therefore $t^4y' +4t^3y=t^3e^{-t}$ $\int \frac{d}{dt}(t^4+y)=\int t^3e^{-t}dt$ $t^4 +y = \int t^3e^{-t}dt$ $y = \int t^3e^{-t}dt -t^3 +C$ apply tabular for int part, he hehe... it's equal but anyway, that's what I have, not yours. either way is the same. but the answer is not in book. hehehe

OpenStudy (anonymous):

sorry i'm not familiar with your 'tabular' method

OpenStudy (anonymous):

so now that you solved for y, you know how to find C now, right? you should be all set!

OpenStudy (anonymous):

Not trying to step on robz8 toes, or anything. Loser66 sent me a private message and asked my input on this problem so let me take a look: $t^3y'+4t^2y=e^{-t}$ To go to standard linear form divide both sides by t^3: $y'+\frac{ 4 }{ t }y=\frac{ e^{-t} }{ t^3 }$ This is a linear equation so we need to find the integrating factor. this is $e^{\int\limits{\frac{ 4 }{ t }}dt}=e^{4\ln(t)}=t^4$ Multiplying through gives the equation: $t^4y'+4t^3y=te^{-t}$ which is $\frac{ d }{ dt }[t^4y]=te^{-t}$ The left side is solved using the FTC. The right side requires integration by parts which gives: $t^4y=-te^{-t}-e^{-t}+C$ So $y= \frac{ e^{-t} }{ t^3 }-\frac{e^{-t}+C}{t^4}$ Substitute the initial conditions y(-1)=0 and solve you'll see that C=0. Thus the solution is: $y= \frac{ e^{-t} }{ t^3 }-\frac{e^{-t}}{t^4}$ You can check on your own that it satisfies both the initial conditions and the differential equation.

OpenStudy (anonymous):

loser66 in your post above where you use an integrating factor, there is a plus sign where there should be a times sign.

OpenStudy (anonymous):

anytime, sorry it taken a while i haven't logged in for a few days. rob was on the right track to helping you

OpenStudy (anonymous):

that formula is just a generalized way to solve the problem and it makes more errors than its worth using sometimes. but you have to use the formula when the right hand side becomes a function you can't integrate