First Order Differential Equations

[Bucky]

Homework Statement

Solve the following differential equation using separation of variables

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

Homework Equations

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?

 

[Dick]

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.

 

[Bucky]

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]

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

 

[Bucky]

ok, right, rearranged it and here's 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 let's 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]

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

 

[Bucky]

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

1/3(1+x)^3

 

[Dick]

You want to integrate 1/(1+x)^2. Try again.

 

[Dick]

Hint: integral of 1/u^2 du=-1/u.

 

[Bucky]

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]

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

 

[Bucky]

ok so that gives...

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

 

[HallsofIvy]

ok, right, rearranged it and here's 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]

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.

 

[Bucky]

i don't see where the missing sign is. unless the sign attatched to the letter comes out? but that's 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.

 

[Dick]

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]

i don't see where the missing sign is. unless the sign attatched to the letter comes out? but that's 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?

 

[Bucky]

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]

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]

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!