Question

Find the solution to the initial value problem
Equations Posted below

Answer 1

\[\frac{ y' - e ^{-t}+6 }{ y } = -6 , y(0) = -5\]

Answer 2

I know you should work with the first eq first and get it into standard form by multiplying both sides by y, then get the y' and y on the same side with the rest on the other side

Answer 3

im stuck from that point on though

Answer 4

i need to find an integrating factor i think... idk

Answer 5

The equation is not separable. It requires the use of an integrating factor. This requires that you "undo" the product rule.

Answer 6

Could you elaborate further

Answer 7

We have:\[\frac{ y' - e ^{-t}+6 }{ y } = -6 , y(0) = -5\]
So we can simplify:
\[y'-e^{-t}+6=-6y\]
And re-arrange:
\[\frac{dy}{dt}+6y=e^{-t}-6\]
And simplify again:
\[\eqalign{
&\frac{dy}{dt}+p(t)y=q(t) \\
where
&\\
&p(t)=6\\
&q(t)=e^{-t}-6}\]
The solution to this since it is a linear DE is:
\[y=\frac{\int{e^{\int{p(t)}dt}q(t)}+c_1}{e^{\int{p(t)}dt}}\]

Answer 8

I am getting \[ \frac{ \frac{ 1 }{ 5}e ^{5t} -e ^{6t} + C}{ e ^{6t} }\]
is that correct?

Answer 9

Sure:

Let's do a few steps:
1)\[\frac{y′−e^{-t}+6}y=−6,\]
Now simplify that:

\[[\frac{y′−e^{−t}+6}y=−6]\rightarrow \frac{ \frac{ d(y)}{dt }}{y}-\frac{ e^{-t} }{ y }+\frac{ 6 }{ y }=6\]

Now we want to get rid of the cumbersome stuff. So, rewrite it again, but this time just move the Y to the other side by multiplication:
\[[\frac{y′−e^{−t}+6}y=−6]\rightarrow \frac{ \frac{ d(y)}{dt }}{1}-\frac{ e^{-t} }{ 1 }+\frac{ 6 }{ 1 }=6y\rightarrow \frac{ d(y)}{dt }- e^{-t} +6=6y\]

Now, that we have a more simple expression:
\[\frac{ d(y)}{dt }- 6y=e^{-t}-6\]

Now we want to get an integrating factor u(x) such that :
\[\mu(x)=e^{\int\limits p(x)dx} \]

Where p(x)=6

Now, multiply the equation by \[\mu(x)\] Such that: \[\mu(x)\frac{dy}{dx}+P(x)\mu(x)=\mu(x)Q(x)\rightarrow\frac{d[\mu(x)y]}{dx}=\mu(x)Q(x)\]

Answer 10

Yes. You are correct.

Answer 11

thanks!