CUET Preparation Today
$(y^3-2x^2y)dx+(2xy^2-x^3)dy=0$ represent the curve: |
$\frac{x}{y}\sqrt{x^2-y^2}=c$ $xy\sqrt{x^2-y^2}=c$ $xy\sqrt{y^2-x^2}=c$ $\frac{x}{y}\sqrt{x^2+y^2}=c$ |
$xy\sqrt{y^2-x^2}=c$ |
Put y = tx; $x\frac{dt}{dx}=\frac{3t(1-t)(1+t)}{2t^2-1};\int(\frac{1}{t}+\frac{1/2}{t-1}+\frac{1/2}{t+1-1})dt+\int\frac{3}{x}dx=const.$ In $t\sqrt{t^2-1}+3\,ln\,x=const.$ $xy\sqrt{y^2-x^2}=c$
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