JEE MAIN - Mathematics Hindi (2024 - 9th April Morning Shift - No. 5)
अवकल समीकरण $$(x^2+y^2) \mathrm{d} x-5 x y \mathrm{~d} y=0, y(1)=0$$ का हल है :
$$\left|x^2-4 y^2\right|^5=x^2$$
$$\left|x^2-2 y^2\right|^6=x$$
$$\left|x^2-2 y^2\right|^5=x^2$$
$$\left|x^2-4 y^2\right|^6=x$$
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