help me in linear differential equation pls

Question

erikaxx · Answer

PROBLEM IS $$(x-2)\frac{ dy }{ dx } = y+2(x-2)^3$$

erikaxx · Answer

@Loser66

erikaxx · Answer

divide the equation by (x-2)

erikaxx · Answer

$$\frac{ dy }{ dx} = \frac{ y }{ x-2 } + 2(x-2)^2$$
$$\frac{ dy }{ dx } - \frac{ y }{ (x-2) } = 2(x-2)^2 $$

erikaxx · Answer

@ganeshie8 check it pls

trevor12 · Answer

im here

trevor12 · Answer

im sorry not good at math

johnweldon1993 · Answer

Good so far...that is the setup you want to work with!

erikaxx · Answer

so the p(x) is - y/(x-2) and q(x) is 2(x-2)^2

johnweldon1993 · Answer

Correct

johnweldon1993 · Answer

Or well, p(x) is actually $\large -\frac{1}{x-2}$ because remember the basic setup is

$$\large \frac{dy}{dx} + p(x)y = q(x)$$

erikaxx · Answer

yes

johnweldon1993 · Answer

So your integration factor is actually $\large e^{\int -\frac{1}{x-2}dx}$

erikaxx · Answer

so e^ln(x-2)

erikaxx · Answer

righr?

erikaxx · Answer

then x- 2 is IF

johnweldon1993 · Answer

Well remember the negative sign in front

$$\large \int -\frac{1}{x-2}dx$$
$\large u = x-2$    $\large du = dx$

$$\large -\int \frac{1}{u}du$$
so we have $\large e^{-ln(u)}$

OR $\large -\frac{1}{u}$ --> $\large \frac{1}{-x + 2}$  is what we are multiplying everything by

erikaxx · Answer

ah sorry forgot the sign

erikaxx · Answer

$$(-\frac{ 1 }{ x^2 })(y) = \int\limits 2(x-2)^2 (-\frac{ 1 }{ x^2 })dx + c$$

johnweldon1993 · Answer

So we need to multiply everything by that integration factor

$$\large \frac{\frac{dy}{dx}}{-x + 2} - \frac{y}{(x-2)(-x + 2)} = \frac{2(x-2)^2}{-x + 2}$$

erikaxx · Answer

/wait.... the formula for the solution is $$(I.F)y = \int\limits Q(x)(IF)dx + c$$

johnweldon1993 · Answer

Oh okay you like doing it that way lol...okay lets write it like that!

in that form, knowing the integration factor is $\large \frac{1}{-x + 2}$

We have $\Large y = \frac{\int \frac{2(x-2)^2}{-x + 2}dx + c}{\frac{1}{-x + 2}}$

Right? *I just divided both sides by the integration factor*

So when we evaluate that integral on top and simplify..what do we get?

erikaxx · Answer

The answer in the book should be $$y = (x-2)^3 + c(x-2)$$

mathmate · Answer

Yes, the book answer is correct.

johnweldon1993 · Answer

Hmm, not what I'm arriving at...Hmm, let me write this all out again and see if I can catch something

$$\large (x - 2)y' = y + 2(x - 2)^3$$
$$\large y' - \frac{y}{x-2} = 2(x - 2)^2$$
Integration factor = $\large e^{\int -\frac{1}{x-2}dx} = e^{-ln(x-2)}=\frac{1}{-x + 2}$
$$\large (\frac{1}{-x+2})y = \int \frac{2(x-2)^2dx}{-x + 2}$$ 

Evaluate out that integral:
$$\large \int \frac{2(x-2)^2}{-x+2}dx = -\int 2(x-2) = -x^2 + 4x + c$$

So we have $\large (\frac{1}{-x+2})y = -x^2 + 4x + c$

or

$$\large y = \frac{-x^2 + 4x + c}{\frac{1}{-x+2}}$$
which is equal to
$$\large (-x^2 + 4x + c)(-x + 2)$$

Hmm, if we expand that out...we get $\large x^3 - 4x^2 + xc - 2x^2 + 8x - 2c$ or $\large (x^3 -6x^2 + 8x) + c(x -2)$  

Hmm...not quite matching...

Where am I going off @mathmate ?

ganeshie8 · Answer

(−x2+4x+c)(−x+2) 

completing the square for the left product should do ?

johnweldon1993 · Answer

For the left hand side?
$$\large x^2 - 4x = c$$
$$\large (x-2)^2 -4 = c$$

ganeshie8 · Answer

(−x2+4x+c)(−x+2)

-((x-2)^2 + C)(−x+2)

johnweldon1993 · Answer

OH I see what you were getting at >.<

mathmate · Answer

I think the algebra would be much simpler if we substitute 
u=(x-2)
which will give
$uy'=y+2u^3$
but the final result would have been the same.

ganeshie8 · Answer

That circus is just to match with the textbook and the OP must see that both the answers are equal up to a constant

johnweldon1993 · Answer

Yeah they're equivalent, just took a little manipulation...good thing too because honestly something so simple as completing the square stumped me haha...thanks guys!