Question

The D.E. whose solution is:
$(x-a{)}^2+(y-b{)}^2=r^2$ where a,b are arbitrary constants

• A. $r^2{y}_2=1+y^2$
• B. $r^2{y}_2^2=1+y_1^2$
• C. $r^2{y}_2^2=[1+y_1^2{]}^4$
• D. $r^2{y}_2^2=[1+y_1^2{]}^3$

Answer 1

The correct option is D $r^2{y}_2^2=[1+y_1^2{]}^3$

Given

$(x-a{)}^2+(y-b{)}^2=r^2----------------------->1$

On differentiating once w.r.t x,

$2(x-a)+2(y-b)\frac{dy}{dx}=0$

$(x-a)+(y-b)\frac{dy}{dx}=0------->2$

On differentiating once again,

$1+(\frac{dy}{dx}{)}^2+(y-b)\frac{d^{2}y}{d{x}^2}=0$

$(y-b)\frac{d^{2}y}{d{x}^2}=-(1+(\frac{dy}{dx}{)}^2)----->3$

From (2)

$(x-a)=-(y-b)\frac{dy}{dx}$

Keeping (x-a) value in (1)

$(y-b{)}^2(\frac{dy}{dx}{)}^2+(y-b{)}^2=r^2$

$(y-b{)}^2((\frac{dy}{dx}{)}^2+1)=r^2--------->4$

Squaring (3) we have

$(y-b{)}^2(\frac{d^{2}y}{d{x}^2}{)}^2=[1+[\left(\frac{dy}{dx}{)}^2\right){]}^2-------->5$

Finally dividing (4/5)

$[1+(\frac{dy}{dx}{)}^2){]}^3=r^2(\frac{d^{2}y}{d{x}^2}{)}^2$

$[1+y_1^2{]}^3=r^2{y}_2^2$(since $\frac{dy}{dx}=y_1,\frac{d^{2}y}{d{x}^2}=y_2$)

Option $D$ is correct