The differential equation corresponding to th | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given the family of circles: $(x-a)^2+(y-b)^2=4$. Differentiating with respect to $x$: $2(x-a)+2(y-b)y'=0 \implies (x-a)+(y-b)y'=0$. Differentiati

Vedclass · Accepted Answer

$4\left(\frac{d^2 y}{d x^2}\right)^2=\left[1+\left(\frac{d y}{d x}\right)^2\right]^3$

Vedclass · Answer

(B) Given the family of circles: $(x-a)^2+(y-b)^2=4$. Differentiating with respect to $x$: $2(x-a)+2(y-b)y'=0 \implies (x-a)+(y-b)y'=0$. Differentiating again with respect to $x$: $1+(y')^2+(y-b)y''=0 \implies (y-b) = -\frac{1+(y')^2}{y''}$. Substituting $(y-b)$ back into the first derivative equation: $(x-a) = -y'(y-b) = y' \cdot \frac{1+(y')^2}{y''}$. Now substitute $(x-a)$ and $(y-b)$ into the original equation: $\left(y' \cdot \frac{1+(y')^2}{y''}\right)^2 + \left(-\frac{1+(y')^2}{y''}\right)^2 = 4$. This simplifies to: $\frac{(1+(y')^2)^2}{(y'')^2} \cdot ((y')^2+1) = 4$. Thus,$(1+(y')^2)^3 = 4(y'')^2$,which is $4\left(\frac{d^2 y}{d x^2}\right)^2 = \left[1+\left(\frac{d y}{d x}\right)^2\right]^3$.

The differential equation corresponding to the family of circles given by $(x-a)^2+(y-b)^2=4$,where $a$ and $b$ are parameters,is

Similar Questions

The differential equation of which $xy = ae^x + be^{-x} + x^2$ is a solution,is

The differential equation whose general solution is given by $y = (c_1 \cos(x + c_2)) - (c_3 e^{(-x + c_4)}) + (c_5 \sin x)$,where $c_1, c_2, c_3, c_4, c_5$ are arbitrary constants,is

$y=A e^x+B e^{2 x}+C e^{3 x}$ satisfies the differential equation

The differential equation of the family of curves for which the length of the normal is equal to a constant $k$,is given by

If $m$ and $n$ are respectively the order and degree of the differential equation of the family of parabolas with focus at the origin and $X$-axis as its axis,then $m n-m+n=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module