Question

Find the integrating factor and use it to find the general solution for the following differential equation:
ty'+y = -5tcos(8t)

Answer 1

@Loser66 @wio
i would have to divide out the t correct

Answer 2

@wio

Answer 3

to me, Yes

Answer 4

so its y'+y = -5cos(8t)

Answer 5

the middle term is \(\dfrac{1}{t} y\)

Answer 6

ok

Answer 7

Look again at the left hand side
\[\Large ty'+y=\frac{d(yt)}{dt} \]

Answer 8

So you're already set :)

Answer 9

wait what...

Answer 10

Finding an integration factor isn't necessary, but if you do, it will turn out to be \(t\) which you need to multiply your DE through in normal standard form, so it will turn out to be the same equation as you have already written up there.

Answer 11

@RichG1, the left side is already the derivative of another expression. All that's left is to integrate.

Answer 12

im so confused

Answer 13

Dividing out by \(t\) yields
\[y'+\frac{1}{t}y=-5\cos8t\]
The integrating factor would be \(\mu(t)=e^{\int1/t~dt}=e^{\ln|t|}=t\).

Multiplying by \(\mu\) yields the same original equation:
\[ty'+y=-5t\cos8t\]
What we're all trying to say is that the integrating factor-finding is not needed here, because you can proceed with solving for \(y\) right away.

Answer 14

so what im getting here is \[e ^{1/t} \int\limits_{} -5\cos8t e ^{1/t}dt +ce ^{1/t}\]

Answer 15

No, the integrating factor was \(t\), not \(e^{1/t}\).

Given
\[ty'+y=-5t\cos8t\]
You have that the left side is derivative of a product:
\[\frac{d}{dt}[ty]=-5t\cos8t\\
ty=-5\int t\cos8t~dt\]

Answer 16

ok then intergrate and devide out t

Answer 17

divide

Answer 18

so -5/8 sin(8 t) + c
?

Answer 19

@SithsAndGiggles

Answer 20

No, the integral wasn't done correctly. First, you have to integrate by parts:
\[\begin{matrix}u=t&&&dv=\cos8t~dt\\du=dt&&&v=\frac{1}{8}\sin8t\end{matrix}\]
\[\begin{align*}-5\int t\cos8t~dt&=-5\left[\frac{1}{8}t\sin8t-\frac{1}{8}\int\sin8t~dt\right]\\
&=-\frac{5}{8}t\sin8t-\frac{5}{16}\cos8t+C
\end{align*}\]
So you have
\[ty=-\frac{5}{8}t\sin8t-\frac{5}{16}\cos8t+C\]

Answer 21

Oh i tried to factor out the t

Answer 22

i dont think this is correct when i factor out the t i get \[\frac{-5 \cos8 t+2 t \sin8t }{ 16 t } \]