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

Question

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

loser66 · Answer

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

anonymous · Answer

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

loser66 · Answer

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

loser66 · Answer

yes

loser66 · Answer

would you please check my y?

anonymous · Answer

give me a sec

anonymous · Answer

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

loser66 · Answer

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

anonymous · Answer

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

loser66 · Answer

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

anonymous · Answer

correct

loser66 · Answer

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

loser66 · Answer

correct?

anonymous · Answer

yes, but what is the need for that?

loser66 · Answer

because of the formula need them to plug in

loser66 · Answer

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

anonymous · Answer

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

loser66 · Answer

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

anonymous · Answer

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

anonymous · Answer

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

loser66 · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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.

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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