The Student Room Group
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>

This discussion is now closed.

Check out other Related discussions

Differential Equation

Hi guys, I'm having trouble with three DE's and I could use some help because I really don't know where to start.

1. dxdy=x3y2\frac{dx}{dy} = x^3y^2

2. x3dydx+3y2=xy2x^3\frac{dy}{dx} + 3y^2 = xy^2

and

3. xdydx=y+x2+y2x\frac{dy}{dx}= y+\sqrt{x^2+y^2}

any help is greatly appreciated, I just don't quite know where to begin with them, do i want all y variables on one side and x's on the other or what. I'm so confused.

Cheers.
Marc
1. It looks like a standard separation-of-variables DE. I'd divide by x³, multiply by dy, then integrate both sides.

2. Integrating factor, I think. Divide everything by x³ to get dy/dx on its own, then you find the integrating factor from what's left on the same side as the dy/dx

3. No idea, but possibly integrating factor (after you rearrange and divide) :o:
Reply 2
Avatar for mdmason
mdmason
OP
cheers thats helped quite a lot with the first two, anybody got any other ideas for the third ?
Reply 3
Avatar for Mathletics
Mathletics
mdmason
cheers thats helped quite a lot with the first two, anybody got any other ideas for the third ?


Try the subtitution y=vx where v is a function of x. That should reduce it to a seperable differential equation in v and x.
Reply 4
Avatar for Salinger
Salinger
1
Mathletics
Try the subtitution y=vx where v is a function of x. That should reduce it to a seperable differential equation in v and x.


This, and remember to substitute dy = vdx + xdv?
Reply 5
Avatar for around
around
13
I think the second is separable as well.
Reply 6
Avatar for mdmason
mdmason
OP
So....

1. dydx=x3y2\frac{dy}{dx} = x^3 y^2
becomes
∫1x3dx=∫y2dy\int\frac{1}{x^3}dx = \int y^2dy
which becomes
−12x2=y33-\frac{1}{2x^2}=\frac{y^3}{3}
so
y3=−32x2y^3=-\frac{3}{2x^2}
and
y=−32x23y = \sqrt[3]{-\frac{3}{2x^2}}


2. x3dydx+3y2=xy2x^3\frac{dy}{dx} + 3y^2 = xy^2
becomes
dydx+1x33y2=y2x2\frac{dy}{dx} + \frac{1}{x^3}3y^2 = \frac{y^2}{x^2}

P(x)=1x3P(x)=\frac{1}{x^3}

so μ=e−12x2\mu = e^{-\frac{1}{2x^2}}

e−12x2dydx+e−12x21x33y2=e−12x2y2x2e^{-\frac{1}{2x^2}}\frac{dy}{dx}+e^{-\frac{1}{2x^2}}\frac{1}{x^3}3y^2=e^{-\frac{1}{2x^2}}\frac{y^2}{x^2}

I get a bit stuck there rearranging for y if anyone can see where I went wrong could you let me know..? cheers.


I'm still working on the third.
Reply 7
Avatar for Mathletics
Mathletics
As my prevous post suggests, you have to make the sub y = vx. Where ever you see a "y", replace it with "vx". You also have to differentiate y = vx so that you have something to sub in for dy/dx.
Reply 8
Avatar for mdmason
mdmason
OP
ok thanks gotcha i think. do the first two look correct above? just want to know that i'm on the right track with them so that i can get the rest of them right in the future
Reply 9
Avatar for Mathletics
Mathletics
mdmason
So....

1. dydx=x3y2\frac{dy}{dx} = x^3 y^2
becomes
∫1x3dx=∫y2dy\int\frac{1}{x^3}dx = \int y^2dy
which becomes
−12x2=y33-\frac{1}{2x^2}=\frac{y^3}{3}
so
y3=−32x2y^3=-\frac{3}{2x^2}
and
y=−32x23y = \sqrt[3]{-\frac{3}{2x^2}}


2. x3dydx+3y2=xy2x^3\frac{dy}{dx} + 3y^2 = xy^2
becomes
dydx+1x33y2=y2x2\frac{dy}{dx} + \frac{1}{x^3}3y^2 = \frac{y^2}{x^2}

P(x)=1x3P(x)=\frac{1}{x^3}

so μ=e−12x2\mu = e^{-\frac{1}{2x^2}}

e−12x2dydx+e−12x21x33y2=e−12x2y2x2e^{-\frac{1}{2x^2}}\frac{dy}{dx}+e^{-\frac{1}{2x^2}}\frac{1}{x^3}3y^2=e^{-\frac{1}{2x^2}}\frac{y^2}{x^2}

I get a bit stuck there rearranging for y if anyone can see where I went wrong could you let me know..? cheers.


I'm still working on the third.


Neither of the first two are correct.

For the first you're given dydx=x3y2\frac{dy}{dx}=x^3y^2 rearranging gives ∫dyy2=∫x3dx\int \frac{dy}{y^2}= \int x^3 dx. I'm sure you can carry on from there.

The second, P(x)= 3x3\frac{3}{x^3}.
Reply 10
Avatar for mdmason
mdmason
OP
oh yeah i see where i went wrong now, cheers.
the second one isn't an integrating factor one (it's not of the correct form for that) , you just need to factorise then seperate the variables.

Related discussions

Latest

Trending

Trending

Articles for you