Question #92847
Find the general solution of the DE
y² lny = x py+p².
Does the equation has any singular solution? If yes, obtain it.
1
Expert's answer
2019-08-20T09:32:36-0400

Let's rewrite the equation in form

lny=xpy+p2y2\ln y=x\frac{p}{y}+\frac{p^2}{y^2}

Denote Y=lnyY=\ln y hence P=dYdx=pyP=\frac{dY}{dx}=\frac{p}{y}

Terefore

Y=xP+P2Y=xP+P^2

which is in the Clairaut's form


The solution can be obtained by just replacing P by constant c.

Y=cx+c2Y=cx+c^2

or

lny=cx+c2\ln y=cx+c^2

y=exp(cx+c2)y=\exp(cx+c^2)


To find a singular solution, let's solve the equation

(P2)+x=0(P^2)'+x=0

2P=x2P=-x

P=x2P=-\frac{x}{{2}}

Therefore

Y=x22+(x2)2=14x2Y=-\frac{x^2}{2}+(-\frac{x}{2})^2=-\frac{1}{4}x^2

will be the singular solution of this equation


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