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Integrating factor, initial value problem


New member
Jan 30, 2013
kxy \frac{dy}{dx} = y^2 - x^2 \quad , \quad
y(1) = 0

My professor suggests substituting P in for y^2, such that:

P = y^2
dP = 2y dy

I am proceeding with an integrating factor method, but unable to use it to separate the variables, may be coming up with the wrong integrating factor ( x )


Indicium Physicus
Staff member
Jan 26, 2012
It's Bernoulli. I would do
kxy \frac{dy}{dx}&=y^{2}-x^{2} \\
\frac{dy}{dx}&= \frac{y}{kx}- \frac{x}{ky} \\
\frac{dy}{dx}- \frac{y}{kx}&=- \frac{x}{ky}.
Then you can see that the appropriate substitution is what your professor suggested, which is $v=y^{1-(-1)}=y^{2}$. Then $dv/dx = 2y dy/dx$. Multiplying the equation by $2y$ yields
2y \frac{dy}{dx}- \frac{2 y^{2}}{kx}&=- \frac{2 x}{k} \\
\frac{dv}{dx}- \frac{2 v}{kx}&=- \frac{2x}{k}.
This is first-order linear in $v$, so you can use the integrating factor method here.