Question

solve bernolli d.e
(xy- (dy/dx))e^x^2=y^3

Answer 1

Is your equation this:
\[ \left( xy - \frac{dy}{dx}\right) e^{x^2} = y^3 \]

Answer 2

rearrange as [2e^(x^2)(y^2)dx]-[2e^(x^2)(ydy)]=2y^4dx
=> 2dx=d{(e^x^2)/(y^2)}
integrating
=> 2x=(e^(x^2))/(y^2) + C.

Answer 3

No, you can't do that because y is a function of x and vice versa.

Answer 4

i think i can. i am just rearranging the differential equation.

Answer 5

the key, I believe, is in getting rid of the y^3 by a useful substition

Answer 6

if u want a simpler solution that just involves working on paper then divide the whole equation by y^3 and then take (y^(-2)) as another variable 't'.
then solve using ur normal method. this way its easier to understand...

Answer 7

\[\left( xy - \frac{dy}{dx}\right) e^{x^2} = y^3\]
\[ \left(e^{x^2}xy - e^{x^2}\frac{dy}{dx} = y^3\right)y^{-3}\]
\[ e^{x^2}xy^{-2} - y^{-3} e^{x^2}\frac{dy}{dx} = 1\]
\[ - y^{-3} e^{x^2} y'+e^{x^2}xy^{-2} = 1\]
ive always subbed a "z" just to make things stand out in my mind
\[z=y^{-2};\ z^{-1/2}=y\to (z^{3/2}=y^{-3});\ -\frac 12 z^{-3/2}z'=y'\]
\[ - y^{-3} e^{x^2} y'+e^{x^2}xy^{-2} = 1\]\[ - e^{x^2}(\cancel{z^{3/2}})(-\frac 12 \cancel{z^{-3/2}}z')+e^{x^2}x\ z = 1\]

\[ \frac{e^{x^2}}{2}\ z'+e^{x^2}x\ z = 1\]

and solve for z stuff

Answer 8

@jamesj - why did u think that rearranging the differential equation was wrong in some way??
it would be helpful if u could explain it in detail...

Answer 9

thnx guys!