Solving differential equation with y'

[ParoXsitiC]

Homework Statement

\left( 1+x \right) ^{2}y' -2\,{\it xy}=0

Homework Equations

The Attempt at a Solution

I know I need to get all the y and dy on one side, all the x and dx on another. Then integrate.
I also know the answer should be y=C(1+x^2)

\left( {x}^{2}+2\,x+1 \right) {\frac {dy}{dx}} =2\,{<br /> \it xy}

\frac {1}{2x} \left( {x}^{2}+2\,x+1 \right) {\frac {dy}{dx}} =\,{<br /> \it y}

 

[Tomer]

So, if you know what to do, where are you stuck?

 

[HallsofIvy]

Homework Statement

\left( 1+x \right) ^{2}y&#039; -2\,{\it xy}=0

Homework Equations

The Attempt at a Solution

I know I need to get all the y and dy on one side, all the x and dx on another. Then integrate.
I also know the answer should be y=C(1+x^2)

\left( {x}^{2}+2\,x+1 \right) {\frac {dy}{dx}} =2\,{<br /> \it xy}

\frac {1}{2x} \left( {x}^{2}+2\,x+1 \right) {\frac {dy}{dx}} =\,{<br /> \it y}

Well, you haven't got "all the y and dy on one side, all the x and dx on another", have you? And, because you want "... dx= ...dy", not 1/dx or 1/dy, I would recommend you get "y and dy" on the left and "x and dx" on the right.

 

[ParoXsitiC]

How would I handle a situation where it is 1/dx and 1/dy ? Just take the inverse of what's there?

so x (1/dx) would turn into (1/x) dx?

Such that:

{\frac {2x{\it}}{ \left( 1+x \right) ^{2}}}dx={\frac {{1}}{y}}dy

 

[ParoXsitiC]

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

2\int \!{\frac {u-1}{{u}^{2}}}{du} = ln |y|2 \int {\frac {1}{u}}{du} - \int{{u}^{-2}}{du} = ln |y|

2ln|u| + \frac{1}{u} + c = ln |y|

y = {e}^{2ln|u|} * {e}^{\frac{1}{{u}}} * {e}^{c}

y = C{u}^{2}{e}^{\frac{1}{{u}}}

y = C{(1+x)}^{2}{e}^{\frac{1}{{(1+x)}}}

Why do I have the extra {e}^{\frac{1}{{(1+x)}}} and mine is 1+x and not 1+x^2??

 

[dynamicsolo]

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

2\int \!{\frac {u-1}{{u}^{2}}}{du} = ln |y|2 \int {\frac {1}{u}}{du} - \int{{u}^{-2}}{du} = ln |y|

Don't forgot to distribute that '2' over both integrals!

I also know the answer should be y=C(1+x^2) .

Indeed? I'm afraid neither y = C(1+x^{2}) nor y = C(1+x)^{2} solves the original DE. Your solution will, once you fix the mistake indicated...

(Sometimes it doesn't pay to know what The Answer is supposed to be, because it isn't...)

 

[CalcStudent]

Don't forgot to distribute that '2' over both integrals!



Indeed? I'm afraid neither y = C(1+x^{2}) nor y = C(1+x)^{2} solves the original DE. Your solution will, once you fix the mistake indicated...

(Sometimes it doesn't pay to know what The Answer is supposed to be, because it isn't...)

The answer is from back of my calculus book...hmm...

 

[dynamicsolo]

Just out of curiosity, which textbook does you use? Be aware that no set of answers in the back of a book is 100%: even in late editions of textbooks, I've found the error rate usually settles to about 1 out of 300 to 400. In first editions, it can be over 1%! (Problem solvers for books are human, too...)

You don't have to take my word for the error here -- try differentiating the book's solution function (or the alternative I suggested) and putting y and y' back into the given differential equation.

 

[ParoXsitiC]

Calculus: Early Transcendental Functions 5e
Larson/Edwards


The question is from 6.2 exercises #9.

9. (1+x^2)y' - 2xy = 0

The solution is listed at the back of the book as:

9. y = C(1+x^2)

I double checked to make sure the problem and solution were right.

 

[dynamicsolo]

For y = C ( 1 + x^{2} ) , the derivative is y&#039; = C \cdot 2x , yes? Putting this into the left-hand side of the differential equation you posted leads to

[ (1 + x )^{2} \cdot C \cdot 2x ] - [ 2x \cdot C ( 1 + x^{2} ) ] .

I don't think that's going to give you zero. (Wouldn't be the first time a solver goofed in a "back-of-the book" answer...)