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  • Please help! A non linear curve has property that the perpendicular distance of the origin from normal at any point of the curve is equal to the distance of point from x - axis . Then the diffrential equation of the curve is

A non linear curve has property that the perpendicular distance of the origin from normal at any point of the curve is equal to the distance of point from x - axis . Then the diffrential equation of the curve is 

  • Option 1)

    Homogeneous 

  • Option 2)

    Non homogeneous 

  • Option 3)

    Bernoulli's form reducible to linear diffrentiable equation form , after substitution 

  • Option 4)

    None of these 

 

Answers (1)

best_answer

As we have learned

Evation of normal at a point (x,y) -

(y-x)= \frac{-1}{\left ( \frac{dy}{dx} \right )}\left ( X-x \right )

- wherein

\frac{dy}{dx} is slop of  tangent at (x,y) on the curve.

 

 Equation of normal is 

(Y-y)= -\frac{dx}{dy}(X-x)

\Rightarrow \frac{dx}{dy} X+Y- (x\frac{dx}{dy}+y)=0

According to question - 

\left | \frac{-(x\frac{dx}{dy}+y)}{\sqrt{(\frac{dx}{dy})^2+1}} \right | = y

SQuaring both sides we get - 

x^2(\frac{dx}{dy})^2+y^2+2xy\frac{dx}{dy}= y^2(\frac{dx}{dy})^2+y^2

\frac{dx}{dy} \left \{ \right.x^2(\frac{dx}{dy})+2xy- y^2(\frac{dx}{dy})\left. \right \}=0

either \frac{dx}{dy}= 0 \: \: or \: \: \frac{dx}{dy}= \frac{2xy}{y^2-x^2}

either \frac{dx}{dy}= 0 \: \: or \: \: \frac{dy}{dx}= \frac{y^2-x^2}{2xy}

2nd equation will give non linear curve and is homogeneous and 1st will give lines x =C 

 

 

 

 

 

 

 


Option 1)

Homogeneous 

Option 2)

Non homogeneous 

Option 3)

Bernoulli's form reducible to linear diffrentiable equation form , after substitution 

Option 4)

None of these 

Posted by

Himanshu

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