# First Order Differential Equations

## Homework Statement

Solve the following differential equation using seperation of variables

$$(1+x)^2 y' = (1-y)^2 , y(1) = 2$$

## The Attempt at a Solution

haven't gotten very far in this at all :/

i've tried dividing both sides by $$(1+x)^2$$, in order to get y' on it's own..

$$y' = \frac{(1-y)^2}{(1+x)^2}$$

but i don't know how to integrate this...but had a go anyway

apparently the rule for integrating an expression in brackets is..

$$\frac{(ax + b)^n}{a(n+1)}$$

so i tried integrating both halves of the fraction seperatley...giving
$$\frac{(-y+1)^3 }{-3y}$$
$$\frac{(x+1)^2}{3x}$$
putting these together and dividing gave

$$\frac{3x(-y+1)^2}{-3y(x+1)^3} + C$$

however i don't think this is accurage, as substituting in 1 and 2 for x and y respectivley gmade C some out as -3/48. Can someone shed some light at where I've went wrong?

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## Answers and Replies

Dick
Science Advisor
Homework Helper
You haven't gotten y on it's own. Write y'=dy/dx and put the dy with the y's and the dx with the x's.

are you saying i need to get the y terms on one side of the equals and the x terms on the other side?

if so then i'm stumped. I'm not sure how to split up the equation in it's initial form.

Dick
Science Advisor
Homework Helper
How about dy/(1-y)^2=dx/(1+x)^2? Does it look split now?

ok, right, rearranged it and heres what i have:

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

$$\frac {1}{(1-y)^3}{3} = {1}{(1+x)^3}{3} + C$$

which, after substituting y(0) = -1 (wrong values on first post BTW) i get

$$\frac{-7}{24} = C$$

I'm pretty sure this is wrong...but anyway lets sub that back into the integrated function

$$\frac{1}{3} (1-y)^3 = \frac{1}{3} (1+x)^3 - 7/24$$
$$\frac 8(1-y)^3 = 8(1+x)^3 - 7$$

the answer in teh book is given as

$$y = \frac{x-1}{1+3x}$$

Dick
Science Advisor
Homework Helper
It would help a lot if you did the integrations correctly. The integral of 1/x^2 is not 1/x^3.

er sorry thats a typo, i get the integral of 1/(1+x)^3 to be....

1/3(1+x)^3

Dick
Science Advisor
Homework Helper
You want to integrate 1/(1+x)^2. Try again.

Dick
Science Advisor
Homework Helper
Hint: integral of 1/u^2 du=-1/u.

so...the integral of 1/(1+x)^2 is just -1 - (1/x) ?

even that doesn't seem right...though my formula book doesn't have the formula for such an instance.

Dick
Science Advisor
Homework Helper
Make a substitution u=x+1. Similarly for the y integral. You have done that, right?

ok so that gives...

$$- \frac{1}{(1-y)} = - \frac{1}{(1+x)}$$

HallsofIvy
Science Advisor
Homework Helper
ok, right, rearranged it and heres what i have:

$$\int \frac{dy}{(1-y)^2} = \int \frac{dx}{(1+x)^2}$$
I added the integral sign on the right side.

$$\frac {1}{(1-y)^3}{3} = {1}{(1+x)^3}{3} + C$$
I think you meant
$$\frac{1}{3(1-y)^3}= \frac{1}{3(1+x)^3}+ C$$
but you've missed a sign. But only one sign. As Dick said you need to substitute for both 1+ x and 1- y. If u= 1- y, what is du?

Dick
Science Advisor
Homework Helper
I think you meant
$$\frac{1}{3(1-y)^3}= \frac{1}{3(1+x)^3}+ C$$
but you've missed a sign. But only one sign. As Dick said you need to substitute for both 1+ x and 1- y. If u= 1- y, what is du?

I think Halls was just cleaning up your notation and doesn't mean to imply that cube is the right power. But you did miss the substitution sign.

i don't see where the missing sign is. unless the sign attatched to the letter comes out? but thats not what i thought you meant when you said

1/u^2 = -1/u

so do you mean that i should have...

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

EDIT: wait that can't be right...yeah i'm lost.

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Dick
Science Advisor
Homework Helper
ok so that gives...

$$- \frac{1}{(1-y)} = - \frac{1}{(1+x)}$$

If you differentiate the right side of this you get 1/(1+x)^2. That's what you want. If you differentiate the left side you get -1/(1-y)^2. Don't forget the chain rule. That's NOT what you want.

HallsofIvy
Science Advisor
Homework Helper
i don't see where the missing sign is. unless the sign attatched to the letter comes out? but thats not what i thought you meant when you said

1/u^2 = -1/u
He certainly never said that! he said that the anti-derivative of 1/u2 is -1/u.

so do you mean that i should have...

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

EDIT: wait that can't be right...yeah i'm lost.
If you let u= 1+ x, then du= dx and dx/(1+x)2= du/u2. What's the anti-derivative of that?

If you let v= 1-y, then dv= -dy and y/(1-y)2= -dv/v2. What's the anti-derivative of that?

ok so i tried a totally different method and it seems to work.

instaed of substitution i tried using the standard integral

$$\int (ax+b)^n = \frac{(ax+b)^{n+1}}{a(n+1)}$$

thanks a lot for your help guys!

Dick
Science Advisor
Homework Helper
ok so i tried a totally different method and it seems to work.

instaed of substitution i tried using the standard integral

$$\int (ax+b)^n = \frac{(ax+b)^{n+1}}{a(n+1)}$$

thanks a lot for your help guys!

Let u=(ax+b), du=a*dx, du/a=dx.

$$\int (ax+b)^n dx= \frac{1}{a} \int (u)^n du = \frac{1}{a} \frac{u^(n+1)}{(n+1)} = \frac{(ax+b)^{n+1}}{a(n+1)}$$

That is a TOTALLY DIFFERENT METHOD!

Dick
Science Advisor
Homework Helper
ok so i tried a totally different method and it seems to work.

instaed of substitution i tried using the standard integral

$$\int (ax+b)^n = \frac{(ax+b)^{n+1}}{a(n+1)}$$

thanks a lot for your help guys!

Let u=(ax+b), du=a*dx, du/a=dx.

$$\int (ax+b)^n dx= \frac{1}{a} \int (u)^n du = \frac{1}{a} \frac{u^{n+1}}{(n+1)} = \frac{(ax+b)^{n+1}}{a(n+1)}$$

That is a TOTALLY DIFFERENT METHOD!