Question

Solve the differential equations
\[\large\rm (1-x^2)\frac{dy}{dx}=x(1-2y+x) \]

\[\large\rm \frac{dy}{dx}+\frac{\sqrt{1+y^2}}{xy(1+x^2} =0\]

Answer 1

@mathmate

Answer 2

the first one looks like a first order linear diff equation
\[(1-x^2)y'=x-2xy+x^2 \\ (1-x^2)y'+2xy=x+x^2 \\ y'+\frac{2x}{1-x^2}y=\frac{x+x^2}{1-x^2}\]

Answer 3

and factor the RHS

Answer 4

the last one looks separable

Answer 5

Yes but we are going to need an integrating factor which is 1/(1-x²). Multiply that on R.H.S and you will end up with a very crude integral

Answer 6

And about the second, I tried that too. Not easy to solve

Answer 7

what crude integral are you talking about

Answer 8

can i see what you have

Answer 9

\[\large\rm ∫\frac{d}{dx} \frac{y}{1-x^2}=∫\frac{x+1}{(1-x^2)(1-x)} \]

Answer 10

you can use partial fractions for the right hand side
also I might have did something wrong but I got
\[\int\limits (\frac{1}{1-x^2}y)' dx=\int\limits \frac{x}{(1-x)(1-x^2)} dx\]

Answer 11

the right hand side there can be rewritten as
\[\int \frac{x}{(1-x)^2(1+x)}dx\]

Answer 12

And about second how should I integrate
\[\large\rm ∫\frac{1}{x(1+x^2)}dx \]

Answer 13

\(\color{#0cbb34}{\text{Originally Posted by}}\) @myininaya
you can use partial fractions for the right hand side
also I might have did something wrong but I got
\[\int\limits (\frac{1}{1-x^2}y)' dx=\int\limits \frac{x}{(1-x)(1-x^2)} dx\]
\(\color{#0cbb34}{\text{End of Quote}}\)
And yes sorry my mistake

Answer 14

do you remember how to do partial fractions?

Answer 15

Oh yes, That didn't came into my mind. I think both of them can be solved by partial fractions. Thank you for help

Answer 16

np
let me know if you get stuck again

Answer 17

\(\color{#0cbb34}{\text{Originally Posted by}}\) @myininaya
np
let me know if you get stuck again
\(\color{#0cbb34}{\text{End of Quote}}\)
Will do. Thanks

Answer 18

Although less pretty maybe, I think you can brute force the first one with power series since it's linear and the second one i think you might be an exact equation but I can't tell by just looking, I'd have to try. Good luck!

Answer 19

\(\color{#0cbb34}{\text{Originally Posted by}}\) @Kainui
Although less pretty maybe, I think you can brute force the first one with power series since it's linear and the second one i think you might be an exact equation but I can't tell by just looking, I'd have to try. Good luck!
\(\color{#0cbb34}{\text{End of Quote}}\)

Answer 20

lmao well you probably aren't expected to use them as methods to solve these then so I won't confuse you more unless you really wanna know and have some time haha

Answer 21

Yes I really wanna know. I like solving things with incomprehensible methods

Answer 22

Perfect! So the first one to use a power series solution what you do is assume the solution can be written as a power series of this form:

\[y = \sum_{n=0}^\infty a_n x^n\]

So you plug this into the differential equation and rearrange the coefficients and all that until the differential equation looks like this:

\[0= \sum_{n=0}^\infty (incomprehensible \ stuff) x^n\]

The incomprehensible stuff will depend solely on \(n\) and so since x is a free variable in order for this to be equal to zero, the incomprehensible stuff must also be equal to zero. It's actually a result of linear algebra, for the same reason a vector is 0 if and only if all the components are 0.

Anywho, so that incomprehensible stuff will get you a recurrence relation which may or may not end up being something you are familiar with, but at any rate it defines a unique solution!

Answer 23

Okay so I am gonna try that. Btw is there a name to this method of solution?

Answer 24

I think power series solution uhh let me find what I believe to be a fairly good resource one moment

Answer 25

http://tutorial.math.lamar.edu/Classes/DE/SeriesSolutions.aspx