Question

I need help solving the initial value problem: dy/dx - (y/x) = xe^(x), y(1) = e - 1

Answer 1

Once again, we have an equation in the form:
\[\frac{dy}{dx}+p(x)y=q(x)\]
In this case:
\[\eqalign{
&p(x)=-\frac{1}{x} \\
&q(x)=x(e)^x \\
}\]

Answer 2

So we can isolate for y:
\[y=\frac{\int{e^{\int{-\frac{1}{x}dx}}x(e)^x}+c_1}{e^{\int{-\frac{1}{x}dx}}}\]

Answer 3

And we arrive to the general equation:
\[y=x(e^x+c_1)\]
Now, we can sub in \(y(1)=e^1-1\)
And derive the specific equation which would be the answer, you tell me if you want it

Answer 4

\[I.F=e ^{\int\limits \frac{ -1 }{x}dx}=e ^{-\ln x}=e ^{\ln x ^{-1}}=x ^{-1}=\frac{ 1 }{ x }\]
\[C.S is y *\frac{ 1 }{ x }=\int\limits x e ^{x}*\frac{ 1 }{ x }dx+c\]
\[\frac{ y }{x }=e ^{x}+c\]

Answer 5

when x=1, y=e-1
use it and find the value of c

Answer 6

Keith, for the last equation above would it be - C_1

Answer 7

?

Answer 8

If I'm to use y(1) = e^1 - 1, aren't I simply plugging x = 1 into the general formula so that it equals the equation: y(1) = e^1 -1. I'm just a little confused when you said derive the specific equation

Answer 9

Ohh well when they say \(y(1)=e^1-1\), That means that there exists the point:
\[\eqalign{
&y=e^1-1 \\
&x=1 \\
}\]
So we can plug this in like so:
\[y=x(e^x+c_1)\]
\[e^1-1=1(e^1+c_1)\]
And solve for c, then plug it back into
\[y=x(e^x+c_1)\]

Answer 10

Anyways, Patrick, I got to go, I will leave the calculation of \(c_1\) to you and good luck haha
If im online, and you would like me to take a look at any question, don't hesitate to give me a shout m'kay?

Answer 11

Ok I see. I actually got the answer correct as its in the book. Thank you for your help

Answer 12

Anytime!