Question

Integrating factor question

Answer 1

Just want to double check with someone that the integrating factor here is e^t^4+c

Answer 2

\[\mu=e^{\int 4t^3\text dx}=e^{t^4}\]

Answer 3

you can drop the constant because it is an Integrating Factor

Answer 4

Cool!

I was just making sure cause the next part of the question has me slightly confused in terms of what to do next, part c to be precise is were I am stuck

Answer 5

i think there is a mistake in (a)

Answer 6

Where abouts?

Answer 7

or i am just confused ,

Answer 8

It worked out for me using u sub, letting u=t^4

Answer 9

In the end I got e^t^4+c

Answer 10

I'm sure you can separate it
\[\frac{dx}{dt}=-4 \left(-t^3+xt^3 \right)\]
\[\frac{dx}{dt}\text{ = }t^3 (-4 x+4)\]
\[\frac{1}{-4x+4} \frac{dx}{dt} \text{ = }t^3\]
Integrate
\[\int\limits \frac{1}{-4x+4} dt\text{ = }\int\limits t^3 \, dt\]

Answer 11

\[-\frac{1}{4} \ln (-4 x+4)\text{ = }\frac{t^4}{4}+c_1\]
Then solve for x(t)
\[x(t)=-\frac{1}{4} e^{-t^4-4 c_1}+1\]

Answer 12

Sorry at the beginning did you just subract both sides by 4t^3?

Answer 13

before factorising

Answer 14

Just multiply -4 into brackets and factor \(t^3\) at the same time

Answer 15

Oh I see, not paying attention when you moved all the terms to the right either!

Answer 16

or
\[x'(t)+4 t^3 x(t)=4 t^3\]
\[x'(t)=-4t^3x(t)+4t^3\]