# problem with first-order nonlinear ordinary differential equation

• Jan 22nd 2010, 04:50 PM
bobey
problem with first-order nonlinear ordinary differential equation
i have problem to find the solution for : $(3x^3y+2xy+y^3)+(x^2+y^2)dy/dx=0$

i have tried the exact equation method :

$(3x^3y+2xy+y^3)dx+(x^2+y^2)dy=0$

thus M(x,y)= $(3x^3y+2xy+y^3)$

and N(x,y)= $(x^2+y^2)$

then deltaM/deltay= $3x^3+2x+3y^2$
and deltaN/deltax= $2x$

Since deltaM/deltay does not equal to deltaN/deltax, this imply that the eqution is not exact

thus, finding/searching for integrating factor :

1. 1/N(deltaM/deltay-deltaN/deltax)= $(3x^3+3y^3)/(3x^3y+2xy+y^3)
$

y cannot be eliminated . thus, this is a function of both x and y, not just x

2. 1/M(deltaN/deltax-deltaM/deltay)= $(3x^3+3y^2)/(x^2+y^2)$

x cannot be eliminated . thus, this is a function of both x and y, not just y

thus i cannot find the integrating factor in order to solve the DE. where i'm gone wrong? can somebody point it out? i guess may be in algebra...

is there anyway for me to solve the de? please help me.... (Crying)
• Jan 22nd 2010, 05:36 PM
kjchauhan
Your equation will be,

$\frac{dy}{dx}=-\frac{3x^3y+2xy+y^3}{x^2+y^2}$

Now take y=vx

Then $\frac{dy}{dx}=v+x\frac{dv}{dx}$

Now substitute it in the above,

you will get,

$\frac{-(1+v^2)dv}{(v^3+3v)}=\frac{(1+x)dx}{x}$

$\frac{-3(1+v^2)dv}{3(v^3+3v)}=\left(\frac{1}{x}+1 \right)dx$

Now integrate....

Result will be,

$e^{3x}(y^3+3x^2y)-1=0$
• Jan 23rd 2010, 08:07 AM
Jester
Quote:

Originally Posted by kjchauhan
Your equation will be,

$\frac{dy}{dx}=-\frac{3x^3y+2xy+y^3}{x^2+y^2}$

Now take y=vx

Then $\frac{dy}{dx}=v+x\frac{dv}{dx}$

Now substitute it in the above,

you will get,

$\frac{-(1+v^2)dv}{(v^3+3v)}=\frac{(1+x)dx}{x}$

$\frac{-3(1+v^2)dv}{3(v^3+3v)}=\left(\frac{1}{x}+1 \right)dx$

Now integrate....

Result will be,

$e^{3x}(y^3+3x^2y)-1=0$

I only got your's to work out if you start with (with the red term)

$\frac{dy}{dx}=-\frac{3x^3y+2xy+{\color{red}{x}} y^3}{x^2+y^2}$