Question

Find all solutions of y'-xy=x^3*y^3.

Answer 1

v=y^-2=1/y^2
y=1/sqrt(v)

-1/(2v^(3/2))*dv/dx-x/sqrt(v)=x^3*1/(v*sqrt(v))

dv/dx+2xv=-2x^3

u(x)=e^x^2

e^x^2*v'+(e^x^2)2xv=-2x^3*e^x^2

Answer 2

@amistre64

Answer 3

@thomaster

Answer 4

@SolomonZelman @ash2326 @Ashleyisakitty @Luigi0210 @ganeshie8

Answer 5

This is a linear differential equation, sorry I have forgotten this :(

Answer 6

That's fine.

Answer 7

Perhaps @amistre64 Can help.

Answer 8

been long removed from me too with little to no practice

Answer 9

all i can possible recollect is to try a reduction formula or seperate

Answer 10

I used the method to solve a Bernoulli equation in the work that I provided above.

Answer 11

burnnoodle is a reduction process yes

Answer 12

there is some y/x method in the bac of my head but i hahvent a clue if we could apply it here

Answer 13

y'-xy=x^3*y^3

y = vx, y' = v+v'x

v +v'x -x(vx)=x^3*(vx)^3

v +v'x -vx^2=v^3 x^6

prolly not

Answer 14

I have never seen that before.

Answer 15

ive prolly never seen it before either, hard to tell with this memory

Answer 16

@SithsAndGiggles

Answer 17

Ahh wait...I do believe you are on the ight track..now my memory is shaky as well but from the:

\[\large (e^x)^2 \times \frac{dv}{dx}+((e^x)^2)2xv=-2x^3(e^x)^2\]
The reverse product rule here would give

\[\large \frac{d}{dx}((e^x)^2v(x)) = -2(e^x)^2x^3\]

And integrate both sides to get

\[\large (e^x)^2v(x) = -(e^x)^2(x^2 + 1)\]

Answer 18

sorry add in the \(\large + c_1\) at the end

Answer 19

That's where I was stucked, how do I integrate -2x^3*e^x^2? I used integration by parts but it only complicated the problem more.

Answer 20

we have a polly that goes to 0, and an exponential that integrates if thats the case

Answer 21

pfft, e^x^2 ... is that ^(x^2)? or (^x)^2

Answer 22

(-2x^3)(e^(x^2))----> I need to integrate this.

Answer 23

I did it as

Answer 24

e^(x^2) does have a taylor polynomial representation

Answer 25

And I integrated using a u-sub first then by parts

\[\large \int -2(e^{x^2})x^3\]

Let u = x^2
du = 2xdx

\[\large - \int e^u udu\]

*udu because 2xdx * x^2 = 2x^3dx

so we have the x^3 taken care of...

now we can do it by parts

Answer 26

Letting

s = u
ds = du
v = e^u
dv = e^udu

So we would have

\[\large -ue^u +\int e^udu\]

And we know the integral there so we would have

\[\large e^u - e^uu + c\]

plug back in x^2 for 'u' and we have

\[\large e^{x^2} - e^{x^2}x^2 + c\]

Factor out a -e^(x^2)

\[\large -e^{x^2}(x^2 - 1) + c\]

Answer 27

god that was a hassle to type >.< lol sorry for the wait

Answer 28

Thanks!