Diff Eq problem: $$\frac{ dy }{ dt } + t^2y = 1$$

So, first I find my integrating factor. $$e ^{\int\limits_{}^{}t^2dt} = e ^{2t+c}$$
Next, I multiply both sides by the integrating factor. $$(\frac{ dy }{ dt } + t^2y)e ^{2t+c} = e ^{2t+c} $$

Then I want to integrate both sides so I should have : $$\int\limits_{}^{}d(e ^{2t+c} )y = \int\limits_{}^{} e ^{2t+c} dt$$

Question

Diff Eq problem: $$\frac{ dy }{ dt } + t^2y = 1$$

So, first I find my integrating factor. $$e ^{\int\limits_{}^{}t^2dt}  = e ^{2t+c}$$ 
Next, I multiply both sides by the integrating factor. $$(\frac{ dy }{ dt } + t^2y)e ^{2t+c} = e ^{2t+c} $$

Then I want to integrate both sides so I should have : $$\int\limits_{}^{}d(e ^{2t+c} )y = \int\limits_{}^{} e ^{2t+c} dt$$

anonymous · Answer

Is this looking good so far @phi  ?

anonymous · Answer

so now I've got $$e ^{2t+c}y = \frac{ 1 }{ 2 }e ^{2t+c} $$

hmmmmm, at this point its not looking right, and not headed toward the answer in the book.

anonymous · Answer

I should have $$y = e ^{-1/3t^3} \int\limits_{}^{}e ^{-1/3t^3}dt+c$$

inkyvoyd · Answer

line 2 is incorrect

inkyvoyd · Answer

$e ^{\int\limits_{}^{}t^2dt}  = e ^{t^3/3+c}$

austinl · Answer

I don't believe you need the "+C" in the exponential.

inkyvoyd · Answer

Austin is correct probably, seeing as you can multiply both sides by whatever expression you want as long as you don't multiply by zero and divide after

austinl · Answer

$y\prime + t^2y=1$

$\mu(t)=e^{\int p(t)dt}$
$p(t)=t^2$
$\int t^2dt=\dfrac{t^3}{3}$

$\mu(t)y\prime+t^2\mu(t)y=\mu(t)$

inkyvoyd · Answer

staph! I wanted the medal

austinl · Answer

Then I do believe that you simply integrate both sides and solve for y if feasible?

anonymous · Answer

Thanks a ton guys!

austinl · Answer

No problemo!