Answer to Question #272452 in Differential Equations for Srikanth

Question #272452

(px+y)^2=py^2

1
Expert's answer
2021-11-30T13:56:11-0500

Solution;

Given;

(px+y)2=px2(px+y)^2=px^2

Let;

xy=vxy=v

Such that;

y+xdydx=dvdxy+x\frac{dy}{dx}=\frac{dv}{dx}

Rewrite as;

y+xp=qy+xp=q

From which;

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

By direct substitution into the equation;

(qyx×x)+y)2=x2(qyx)(\frac{q-y}{x}×x)+y)^2=x^2(\frac{q-y}{x})

Resolve to;

q2=xqxyq^2=xq-xy

q2=xqvq^2=xq-v

Rewrite ;

v=xqq2v=xq-q^2

Which is an equation of the form;

y=qx+f(p)y=qx+f(p)

In which q=constant

Therefore;

v=cxc2v=cx-c^2

But v=xy;

xy=xcc2xy=xc-c^2

Is the solution.




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