Differential Equations

[delicate crag]

I need help with these questions I have posted my attempt below but I can't see any way out.

[gilded sleet]

for 1. $\frac{dy}{dx}=b+2cx+3dx^{2}$

[fading fjord]

Helcovich Emire

[gilded sleet]

but you're probably looking to write it in terms of y as well, not just x

[delicate crag]

So number 1 the answer is $y^(4)=0$ since that is the DE for that general solution right?

[fading fjord]

codo0160

[gilded sleet]

y(4)=0 is not a differential equation

you need a dy/dx term for it to be a differential equation

[delicate crag]

I meant to write the 4th derivative

[gilded sleet]

you want a fourth order equation?

that'd work

the second one

simplifies to $x^{2}-2xc+y^{2}=0$

[fading fjord]

Helcovich Emire

[gilded sleet]

what does it mean to eliminate the constant

can I like see more context to the problems

like maybe what the assignment is called, or any overarching directions

i'm confused by what you are trying to do

in your work

@delicate crag

[delicate crag]

It's about solutions to differential equations

It seems the only way to eliminate the constant or the terms they are asking you to is to differentiate it so that a multiplied by zero

[delicate crag]

But dy/dx can be written as y' and d2y/dx2 as y'' and so on

[gilded sleet]

well yes

but what you had was y(4)

[delicate crag]

Yes and in y(4) there was no constant all were eliminated

[gilded sleet]

yes

I just didn't know you were suppose to eliminate all the constants in the first one

[delicate crag]

Oh

[gilded sleet]

so for the second one youre suppose to make a differential equation that pdocues that general equation

and the differnetial equation can't have the c in it?

[delicate crag]

I think so

But I'm not sure that question is confusing me rn

[gilded sleet]

and it can be any order?

[delicate crag]

Yes

[gilded sleet]

so like it can have the first derivative, or second derivative or third, etc

oh wait i know

rewrite c in terms of y and x

so the equation is $y_{1}=\sqrt{2xc-x^{2}}$

and the differentrial equation for it is $\frac{dy}{dx}=\frac{c-x}{y}$

[fading fjord]

Helcovich Emire

Helcovich Emire

[delicate crag]

I'm getting

[gilded sleet]

use the first equation $x^{2}-2xc+y^{2}=0$ to solve for c in terms of x and y

[fading fjord]

Helcovich Emire

[delicate crag]

$y'=\frac{c-x} {y} $

[gilded sleet]

'so c=$\frac{x^{2}+y^{2}}{2x}$

[fading fjord]

Helcovich Emire

[gilded sleet]

i got the equation, did you?

[delicate crag]

Did it become 0=0?

[gilded sleet]

plug that equation in for c

0=0 is not a differnetial equation

[gilded sleet]

here

[gilded sleet]

plug for c into this equation

[gilded sleet]

you plugged back for c into the original equation

well ofc you'd get 0=0

if you plug for c into the equation you used to solve for c

plug for c into the differential equation

[delicate crag]

$yy'=\frac{x^2+y^2}{2x}-x$

[fading fjord]

codo0160

[delicate crag]

This?

[gilded sleet]

why is y being multiplied to y'?

[delicate crag]

Normal y

[gilded sleet]

oh i see what you did

it is correct technically

but you always want to have y' by itself

[delicate crag]

Oh oh ok

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

[fading fjord]

codo0160

[delicate crag]

Sorry all divided by y(the rhs)

[gilded sleet]

it simplifies a bit more

[delicate crag]

Oh ok

[gilded sleet]

hint: get rid of the nested fractions

[gilded sleet]

combine the like terms next

[delicate crag]

$y'=\frac{y^2-x^2}{2xy}$

[fading fjord]

codo0160

[gilded sleet]

yes

so for number 3

is x a function of both w and t?

and B and a are the constants?

[delicate crag]

I think t only

[gilded sleet]

it says don't eliminate the w

a parameter is a "control" variable, because you can control what value it is

which you treat just like another independent variable

[delicate crag]

Oh ok but do we differentiate wrt to t right since t is the independent variable?

[gilded sleet]

oh yes

[delicate crag]

So then my solution in the description is right?

[gilded sleet]

i always switch to y and x where y is the function of x

i just find it confusing with x being the dependent variabe

[delicate crag]

Oh oh ok

As in $y''=-w^2y$.?

[fading fjord]

codo0160

[gilded sleet]

yes your work looks correct

[delicate crag]

OK thank you very much

[delicate crag]

.close