Answer:
Step-by-step explanation:
[tex]\frac{y^2}{2x^2} +\frac{4y}{x} +18 ln(\frac{y-5x}{x} )=ln x+42\\diff.~w.r.t.~x\\\frac{1}{2} \frac{x^2 *2y \frac{dy}{dx} -y^2*2x}{x^4} +4\frac{x\frac{dy}{dx} -y*1}{x^2} +18*\frac{x}{y-5x} *\frac{x(\frac{dy}{dx} -5)-(y-5x)*1}{x^2} =\frac{1}{x} +0\\or \frac{xy\frac{dy}{dx} -y^2}{x^3} +\frac{4x\frac{dy}{dx} -4y}{x^2} +\frac{18\frac{dy}{dx} -90}{y-5x} -\frac{18}{x} -\frac{1}{x} =0\\\\multiply~by~x^3(y-5x)\\(y-5x)(xy\frac{dy}{dx} -y^2)+x(y-5x)(4x\frac{dy}{dx} -4y)+x^3(18\frac{dy}{dx} -90)-19x^2(y-5x)=0\\[/tex]
[tex](xy^2-5x^2y)\frac{dy}{dx} -y^3+5xy^2+(4x^2y-20x^3)\frac{dy}{dx} -4xy^2+20x^2y+18x^3\frac{dy}{dx} -90x^3-19x^2y+95x^3=0\\(xy^2-5x^2y+4x^2y-20x^3+18x^3)\frac{dy}{dx} =y^3-5xy^2+4xy^2-20x^2y+90x^3+19x^2y-95x^3\\(xy^2-x^2y-2x^3)\frac{dy}{dx} =y^3-xy^2-x^2y-5x^3\\\frac{dy}{dx} =\frac{y^3-xy^2-x^2y-5x^3}{xy^2-x^2y-2x^3}[/tex]
Hence the result.
