If the independent variable x is changed to y- then the differential equation xd2ydx2 - dydx3- dydx- 0 is changed to xd2xdy2 - dxdy2- k where k equals

Dydx=1dxdy; d^2ydx^2=ddy ( 1dxdy ).dydx= 1 ( dxdy )^3d^2 xdy^2 Hence, x d^2 ydx^2+ ( dydx )^3 yddx=0 becomes x. 1 ( dxdy )^3d^2xdy^2+1 ( dxdy )^3 1 ( dxdy )=0 ...

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Standard XII

Mathematics

Geometrical Applications of Differential Equations

If the indepe...

Question

If the independent variable $x$ is changed to $y$ , then the differential equation $x \frac{d^{2} y}{d x^{2}} + {(\frac{d y}{d x})}^{3} - (\frac{d y}{d x}) = 0$ is changed to $x \frac{d^{2} x}{d y^{2}} + {(\frac{d x}{d y})}^{2} = k$ where $k$ equals

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Solution

$\frac{d y}{d x} = \frac{1}{\frac{d x}{d y}}; \frac{d^{2} y}{d x^{2}} = \frac{d}{d y} ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ \frac{1}{\frac{d x}{d y}} ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ . \frac{d y}{d x} = - \frac{1}{{(\frac{d x}{d y})}^{3}} \frac{d^{2} x}{d y^{2}}$
Hence, $x \frac{d^{2} y}{d x^{2}} + {(\frac{d y}{d x})}^{3} - \frac{y d}{d x} = 0$
becomes $- x . \frac{1}{{(\frac{d x}{d y})}^{3}} \frac{d^{2} x}{d y^{2}} + \frac{1}{{(\frac{d x}{d y})}^{3}} - \frac{1}{(\frac{d x}{d y})} = 0$
or $x \frac{d^{2} x}{d y^{2}} - 1 + {(\frac{d x}{d y})}^{2} = 0$ or $x \frac{d^{2} x}{d y^{2}} + {(\frac{d x}{d y})}^{2} = 1$
$∴ k = 1$

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Q. If the independent variable

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is changed to

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x \frac{d^{2} y}{d x^{2}} + {(\frac{d y}{d x})}^{3} - (\frac{d y}{d x}) = 0

is changed to

x \frac{d^{2} x}{d y^{2}} + {(\frac{d x}{d y})}^{2} = k

where

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If the independent variable x is changed to y, then the differential equation xd2ydx2+(dydx)3−(dydx)=0 is changed to xd2xdy2+(dxdy)2=k where k equals

dydx=1dxdy;d2ydx2=ddy⎛⎜ ⎜ ⎜ ⎜⎝1dxdy⎞⎟ ⎟ ⎟ ⎟⎠.dydx=−1(dxdy)3d2xdy2 Hence, xd2ydx2+(dydx)3−yddx=0 becomes −x.1(dxdy)3d2xdy2+1(dxdy)3−1(dxdy)=0 or xd2xdy2−1+(dxdy)2=0 or xd2xdy2+(dxdy)2=1 ∴k=1

If the independent variable $x$ is changed to $y$ , then the differential equation $x \frac{d^{2} y}{d x^{2}} + {(\frac{d y}{d x})}^{3} - (\frac{d y}{d x}) = 0$ is changed to $x \frac{d^{2} x}{d y^{2}} + {(\frac{d x}{d y})}^{2} = k$ where $k$ equals