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$A curve passes through (2, 3) and satisfying the D.E \int_{0}^{x} ty (t) d t = x^{2} y (x), x > 0 is$

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x2+y2=13

y2=92x

x28+y218=1

xy=C

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detailed solution

Correct option is D

∫0x ty(t)dt=x2y(x) Differentiating w.r.t. x, we getxy(x)+x2y1(x)=0→xdydx+y=0⇒log⁡x+log⁡y=log⁡C⇒xy=C

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A curve passes through (2,3) and satisfying the D.E ∫0x ty⁡(t)dt=x2y(x),x>0 is

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$A curve passes through (2, 3) and satisfying the D.E \int_{0}^{x} ty (t) d t = x^{2} y (x), x > 0 is$