Answer to Question #272452 in Differential Equations for Srikanth

Question

(px+y)^2=py^2

Answer 1

Solution;

Given;

$(px+y)^2=px^2$

Let;

$xy=v$

Such that;

$y+x\frac{dy}{dx}=\frac{dv}{dx}$

Rewrite as;

$y+xp=q$

From which;

$p=\frac{q-y}{x}$

By direct substitution into the equation;

$(\frac{q-y}{x}×x)+y)^2=x^2(\frac{q-y}{x})$

Resolve to;

$q^2=xq-xy$

$q^2=xq-v$

Rewrite ;

$v=xq-q^2$

Which is an equation of the form;

$y=qx+f(p)$

In which q=constant

Therefore;

$v=cx-c^2$

But v=xy;

$xy=xc-c^2$

Is the solution.