ODE problem, stuck!

ironictoaster

Currently having problems with this question.


First thing I did was rearrange the equation z=y/x to y=zx. I got y' using the product rule and got y' = dz/dx*x + z

I subbed y' into the original equation and played around with it to see if I could make it work. I tried factorising and rearranging and I can't get it seem to work out.

Any ideas?

ironictoaster

Nevermind, figured it out, wolfram put me in the right direction.

Michael Collins

If you get a chance maybe you could outline your solution here? Just in case someone comes across this problem/thread in the future.

sponsoredwalk

In terms of methods this equation has the form of a Bernoullli differential equation:

[latex] xyy' = 3x^3 e^x + xe^x y^2 + y^2[/latex]
[latex] xyy' = 3x^3 e^x + (xe^x + 1)y^2[/latex]
[latex] y' = (3x^2 e^x)y^{-1} + (e^x + \frac{1}{x})y[/latex]
[latex] y' = P(x)y + Q(x)y^{-1}[/latex]

Which on substituting z = y², differentiating it w.r.t. x & replacing terms in

[latex] yy' = P(x)y^2 + Q(x)[/latex]

reduces it to a linear differential equation. But following the recommended substitution we find

[latex] z(x) = \frac{y}{x}[/latex]
[latex] z'(x) = \frac{y'}{x} - \frac{y}{x^2}[/latex]
[latex] z'(x) = \frac{y'}{x} - \frac{z}{x}[/latex]
[latex] zz'(x) = \frac{zy'}{x} - \frac{z^2}{x}[/latex]
[latex] zz'(x) = \frac{1}{x}(zy' - z^2)[/latex]
[latex] zz'(x) = \frac{1}{x}( \frac{y}{x}[(3x^2 e^x)y^{-1} + (e^x + \frac{1}{x})y] - z^2)[/latex]
[latex] zz'(x) = \frac{1}{x}( \frac{y}{x}[(3x^2 e^x)y^{-1} + (e^x + \frac{1}{x})y] - z^2)[/latex]
[latex] zz'(x) = \frac{y}{x^2}[(3x^2 e^x)y^{-1} + (e^x + \frac{1}{x})y] - \frac{1}{x}z^2[/latex]

[latex] zz'(x) = \frac{1}{x^2}[(3x^2 e^x) + (e^x + \frac{1}{x})y^2] - \frac{1}{x}z^2[/latex]

[latex] zz'(x) = [ \frac{1}{x^2}(3x^2 e^x) + \frac{1}{x^2}(e^x + \frac{1}{x})y^2] - \frac{1}{x}z^2[/latex]

[latex] zz'(x) = [ (3e^x) + (e^x \frac{y^2}{x^2} + \frac{y^2}{x^3})] - \frac{1}{x}z^2[/latex]

[latex] zz'(x) = (3e^x) + (e^x \frac{y^2}{x^2} + \frac{y^2}{x^3}) - \frac{y^2}{x^3}[/latex]

[latex] zz'(x) = 3e^x + e^x z^2 [/latex]

[latex] zz'(x) = e^x(3 + z^2) [/latex]

which seems pretty ridiculous substitution since it still takes the form of a Bernoulli equation but maybe evaluating the original is too complicated when you explicitly write out the linear part idk...