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  • Find the general solution of y square dx plus x square minus xy plus y square dy equal to 0

Find the general solution of y^{2}dx + (x^{2} - xy + y^{2}) dy = 0.

Answers (1)

Given:

y^{2}dx+(x^{2}-xy+y^{2})dy=0

To find: Solution for the given differential equation

Rewrite the given equation

y^{2}\frac{dx}{dy}=xy-x^{2}-y^{2}\\ \frac{dx}{dy}=\frac{x}{y}-1-\frac{x^{2}}{y^{2}}

It is a homogenous differential equation

Assume x=vy

Differentiating on both the sides

\frac{dx}{dy}=v+y\frac{dv}{dy}

Substitute dy/dx in the given equation

v+y\frac{dv}{dy}=\frac{x}{y}-1-\frac{x^{2}}{y^{2}}

Substitute v=x/y

v+y\frac{dv}{dy}=v-1-v^{2}\\ y\frac{dv}{dy}=-1-v^{2}\\\\ \frac{dv}{1+v^{2}}=-\frac{dy}{y}

Integrating on both the sides

\int \frac{dv}{1+v^{2}}=-\int \frac{dy}{y}\\ \tan^{-1}v=-\ln\;y+c\\ Formula: \int \frac{dx}{x}=\ln x +c\\\int \frac{dv}{1+v^{2}} =\tan^{-1}v+c\\ Substituting \; v=\frac{x}{y}\\ \tan^{-1}\frac{x}{y}=-\ln y+c

Posted by

infoexpert24

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