Is anybody in here smart enough to solve this?

# Is anybody in here smart enough to solve this?

I'm assuming y' is d(y)/dy. Nonetheless, you have 2 unknowns and 1 equation. This cannot be solved without another given linearly independent equation.

solution is y(t) = c_1 e^(-t^2) + 2 sqrt(π) e^(-t^2) erfi(t)

btw.

pls don't make yourself look retarded and ask me what erfi is

y' = 4 - 2ty

y = 4y + ty^2

1 = 4 + ty

-3 = ty

t = -3/y

y' + 2ty = 4

4 - 2ty + 2ty = 4

4 - 2y(-3/y) + 2y(-3/y) = 4

-2y(-3/y) + 2y(-3/y) = 0

6 - 6 = 0

what do i win

Wow it's almost like t is a variable not any random constant you make up

I'm actually a freshman LOL I'm super smart though I took AP calc in high school and got a 3 on the test

I'm super smart

lol

anyone who says this unironically is 100% a fucking brainlet

then pls tell us what the correct solution was, it's already been posted here.

Let's see who the brainlet is

I hate this way of writing. If y is a function, and t is the variable, you should write

[math]\forall t, 3y'(t) + 6ty(t) = 12[/math]

You wouldn't have stupid answers as

But somehow mathematicians find this rigorous...

First order linear ODE?

I could do that in high school.

Integrating factors are for babies.

no, anyone who knows differential equations would know what OP means. they are just retarded highschoolers who try to look smart.

3 unknowns* (also y'). Also needs an initial condition (another lin indep equation).

[math]

3y' + 6ty = 12 \\

3\dfrac{dy}{dt} + 6ty = 12 \\

y=4t \\

\dfrac{dy}{dt} = 4\\

t=0

[/math]

Not him but our school accidentally gave us the bc instead of the ab. Still passed lul.

didn't even know sequences and series other than how to do a taylor expansion and got a 5 on the BC exam

it was definitely the easiest AP test i took

This is an embarrassing performance by all the anons posting before me.

In any case, here we have an inhomogeneous linear first order ODE. This would indicate that we should use an integrating factor. First, divide by 3:

y' + 2ty = 4

Then, we will try the integrating factor u(t) = e^(integral of 2t) = e^(t^2).

Multiplying through, we get

y'*e^(t^2) + y*2te^(t^2) = 4e^(t^2)

The left hand side is just

(y*e^(t^2))' = 4e^(t^2)

Now we integrate:

y*e^(t^2) + C = integral of 4e^(t^2) (which sadly has not got a closed form)

so if y is a solution, it must be of the form y = (C + integral of 4e^(t^2))*e^(-t^2)

Using initial conditions, C can be determined.

Then, you should plug the answer back in to show that y is indeed a solution (fuck doing that though).