Question

If the curve y = f(x) passing through the point (1, 2) and satisfies the differential equation xdy + (y + x^3y^2)dx = 0, then

Answer 1

$\underline{\textsf{Given:}}$

$\textsf{Differential equation is}$

$\mathsf{x\,dy+(y+x^3y^2)dx=0}$

$\underline{\textsf{To find:}}$

$\textsf{Equation of the curve passses through (1,2) and satisfies given}$

$\mathsf{differential\;equation}$

$\underline{\textsf{Solution:}}$

$\mathsf{Consider,}$

$\mathsf{x\,dy+(y+x^3y^2)dx=0}$

$\mathsf{x\,dy+y\,dx+x^3y^2\,dx=0}$

$\mathsf{x\,dy+y\,dx=-x^3y^2\,dx}$

$\mathsf{x\,dy+y\,dx=-(xy)^2\,xdx}$.....(1)

$\bf\,Take,\;t=xy$

$\bf\dfrac{dt}{dx}=x\dfrac{dy}{dx}+y.1$

$\bf\,dt=x\,dy+y\,dx$

$\mathsf{Now,\;(1)\,becomes}$

$\mathsf{dt=-t^2\,xdx}$

$\mathsf{t^{-2}dt=xdx}$

$\mathsf{Integrating}$

$\mathsf{\int\,t^{-2}dt=\int\,xdx}$

$\mathsf{\dfrac{t^{-1}}{-1}=\dfrac{x^2}{2}+C}$

$\mathsf{\dfrac{-1}{t}=\dfrac{x^2}{2}+C}$

$\mathsf{\dfrac{-1}{xy}=\dfrac{x^2}{2}+C}$

$\textsf{Since the curve passes through (1,2)}$

$\mathsf{\dfrac{-1}{2}=\dfrac{1}{2}+C}$

$\mathsf{-\dfrac{1}{2}-\dfrac{1}{2}=C}$

$\mathsf{C=-1}$

$\therefore\mathsf{The\;required\;curve\;is}$

$\mathsf{\dfrac{-1}{xy}=\dfrac{x^2}{2}-1}$

$\mathsf{\dfrac{-1}{xy}-\dfrac{x^2}{2}=-1}$

$\mathsf{\dfrac{2+x^3y}{2xy}=1}$

$\boxed{\mathsf{x^3y-2xy-2=0}}$

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