2
$\begingroup$

I have the following:

$ \dfrac{dx}{dz} = \dfrac{x(x+y)}{(y-x)(2x+2y+z(x,y))} $

I've tried to solve through using an integrating factor, i.e. I rearranged to get:

$ \dfrac{dz(x,y)}{dx} - \dfrac{(y-x)}{x(x+y)}z(x,y) = \dfrac{2(y-x)}{x} $

and used IF $= \dfrac{(x+y)^2}{x} $. I do get an eventual (fairly messy, but ok) answer. I would appreciate any feedback on the validity of the method - in other words, because $z$ is a function of $x$ and $y$, is it 'acceptable' to rearrange it in this way and then use integrating factor.

Help/feedback very appreciated.

2 Answers 2

1

You can solve the differential equation by considering $ \frac{dz}{dx} $ instead of $ \frac{dx}{dz} $. Given,

$\dfrac{dx}{dz} = \dfrac{x(x+y)}{(y-x)(2x+2y+z(x,y))} $

$ \implies \dfrac{dz}{dx} = \dfrac{(y-x)(2x+2y+z(x,y))}{x(x+y)}=\frac{(y-x)}{x(x+y)}z + \frac{2(y-x)}{x} . $

$ \implies \dfrac{dz}{dx} = \frac{(y-x)}{x(x+y)}z + \frac{2(y-x)}{x}. $

Now, the above is a linear first order differential equation which can be solved using standard techniques. Keeping in mind that the constant is a function in $y$, the solution is

$ z(x,y) = \frac{-{x}^{3}-2\,y\,x^2-2\,{{{y}^{3}}}+2\,x\,{y}^{2}\ln(x)+x\,g(y)}{(x+y)^2} $

  • 0
    @Sarah24: You are welcome. Glad to assist. Keep the hard work.2012-12-11
1

It would be better in this context either to use partial derivative sign, or suppress $y$ as an argument and treat it purely as a parameter. To cross-check your solution a calculation by means of variating the constant of integration can be carried out. First solve the homogeneous equation $\frac{dz}{z}=\frac{y-x}{x\left(x+y\right)}dx$ $\frac{dz}{z}=\frac{x+y-2x}{x\left(x+y\right)}dx$ $\frac{dz}{z}=\frac{dx}{x}-2\frac{d\left(x+y\right)}{x+y}$ $z=\frac{C(y)x}{\left(x+y\right)^{2}}\tag{1}$ Now variate $C$ in $x$ $C=C\left(x,y\right)$ $z_x=\frac{C_xx}{\left(x+y\right)^{2}}+C\left[\frac{1}{\left(x+y\right)^{2}}-\frac{2x}{\left(x+y\right)^{3}}\right]=\\\frac{C_x\left(x\right)x}{\left(x+y\right)^{2}}+C\frac{y-x}{\left(x+y\right)^{3}}$ Inserting the results in the original equation: $ \frac{C_xx}{\left(x+y\right)^{2}}+C\frac{y-x}{\left(x+y\right)^{3}}-\frac{y-x}{x\left(x+y\right)}\frac{Cx}{\left(x+y\right)^{2}}=\frac{2\left(y-x\right)}{x}$ $\frac{C_xx}{\left(x+y\right)^{2}}=\frac{2\left(y-x\right)}{x}$ Solving for $C_x$: $C_x=\frac{2\left(x+y\right)^{2}\left(y-x\right)}{x^{2}}=2\frac{\left(x^{2}+2xy+y^{2}\right)\left(y-x\right)}{x^{2}}=2\frac{x^{2}y+2xy^{2}+y^{3}-x^{3}-2x^{2}y-xy^{2}}{x^{2}}=2\frac{-x^{3}-yx^{2}+xy^{2}+y^{3}}{x^{2}}$ $C_x=2\left(-x-y+\frac{y^{2}}{x}+\frac{y^{3}}{x^{2}}\right)$ Now integrate with respect to $x$: $C=2\left(-\frac{x^{2}}{2}-xy+y^{2}\ln x-\frac{y^{3}}{x}\right)+C_0(y)$ Substitute in $(1)$ to get the final result.

  • 0
    Thank-you, also a useful answer.2012-12-11