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 >>

Differential equation question

solve the differential equation: xy dy/dx - y^2 = (3x^2)(e^(2y/x))
Original post by zomgleh
solve the differential equation: xy dy/dx - y^2 = (3x^2)(e^(2y/x))

Post your working so that appropriate hints can be given. If you can't see how to start, the obvious first step would be to divide by xyxy.
Reply 2
Avatar for ztibor
ztibor
10
Original post by zomgleh
solve the differential equation: xy dy/dx - y^2 = (3x^2)(e^(2y/x))


Take out y as factor at LHS and divide by x^2
yx2⋅(xdydx−y)=3e2yx\displaystyle \frac{y}{x^2}\cdot \left (x\frac{dy}{dx}-y\right )=3e^{2\frac{y}{x}}
With substitution of
t=yx\displaystyle t=\frac{y}{x}
so
y=t⋅x\displaystyle y=t\cdot x
and
dtdx=xdydx−yx2\displaystyle \frac{dt}{dx}=\frac{x\frac{dy}{dx}-y}{x^2}
The equation will be separable
t⋅xdtdx=3e2t\displaystyle t\cdot x \frac{dt}{dx}=3e^{2t}
Reply 3
Avatar for zomgleh
zomgleh
OP
Original post by Farhan.Hanif93
Post your working so that appropriate hints can be given. If you can't see how to start, the obvious first step would be to divide by xyxy.


Okay, so I get dy/dx = 3xe^(2y/x) + (y/x)

Am I right to say this equation is homogeneous of degree 0? The right hand side works out fine but I'm not sure about the dy/dx?
Original post by zomgleh
Okay, so I get dy/dx = 3xe^(2y/x) + (y/x). Am I right to say this equation is homogeneous of degree 0?

It will be if you manage to work out the RHS correctly. :P

The right hand side works out fine but I'm not sure about the dy/dx?

I'm not sure what you mean by "not sure about the dy/dx". A first order ODE is said to be homogeneous if dy/dx can be expressed as some function of (y/x); which, currently, it cannot (but will be possible if you fix the aforementioned error).
(edited 11 years ago)

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you