t*y'-2*y=t^5*sin(2*t)-t^3+4*t^4

(PEMDAS notation)

It's a differential equation. What's its solution?

Let's see how good you guys are. :P

Question

t*y'-2*y=t^5*sin(2*t)-t^3+4*t^4

(PEMDAS notation)

It's a differential equation. What's its solution?

Let's see how good you guys are. :P

across · Answer

$$ty'-2y=t^5\sin(2t)-t^3+4t^4$$

anonymous · Answer

I assume you know how to do it?

anonymous · Answer

across · Answer

I just want to corroborate my answer.

anonymous · Answer

Well Wolfram does it

anonymous · Answer

It seem little too advanced for me. I just started Fourier series

across · Answer

This is what I dribbled down to:

$$y(t)=-(1/2)t^4\cos(2t)+(1/2)t^3\sin(2t)+1/4t^2\cos(2t)-t^3+2t^4+ct^2$$

anonymous · Answer

Sorry ...I am not that Good ..but I will bring you some Help

anonymous · Answer

Lemme see if I can work it out. It's first order linear ODE. Should be solvable using the integrating factor..

across · Answer

Yes, that's correct. I forgot to mention it's a linear differential equation. I used the integrating factor $$\mu(t)=t^{-2}$$.

anonymous · Answer

Yep, that's what I have for the integrating factor also

anonymous · Answer

Yeah, I have what you got so far for y(t)

across · Answer

Thank goodness :)

I just wanted to make sure I didn't mess up somewhere in between my calculations.

Thank you! <3

anonymous · Answer

Nope, looks good. Messy, but I think it's right.