Differentiation (Part-38) | S N De

 

$\,18.\,$ Find a differential equation which is satisfied by all the curves $\,y=Ae^{2x}+Be^{-x/2},\,$ where $\,A\,$ and $\,B\,$ are non-zero constants.

Sol.  $~~~~~~y=Ae^{2x}+Be^{-x/2} \\ \Rightarrow ye^{x/2}=Ae^{5x/2}+B \\ \therefore \frac 12e^{x/2}y+ e^{x/2}\frac{dy}{dx}=\frac{5A}{2}e^{5x/2} \\ \Rightarrow e^{(x/2)-(5x/2)}y+2e^{(x/2)-(5x/2)}\frac{dy}{dx}=5A \\ \Rightarrow e^{-2x}y+2e^{-2x}\frac{dy}{dx}=5A \\ \therefore -2e^{-2x}y+e^{-2x}\frac{dy}{dx}-4e^{-2x}\frac{dy}{dx}\\~~~~~~+2e^{-2x}\frac{d^2y}{dx^2}=0 \\ \Rightarrow 2e^{-2x} \frac{d^2y}{dx^2}-3e^{-2x}\frac{dy}{dx}-2e^{-2x}y=0 \\ \therefore 2\frac{d^2y}{dx^2}-3\frac{dy}{dx}-2y=0~~~\text{(ans.)}$

$\,19.\,$ Form the differential equation of the family of hyperbolas $\,b^2x^2-a^2y^2=a^2b^2\,$ by eliminating arbitrary constants $\,a\,$ and $\,b.$

Sol.  $~~~~b^2x^2-a^2y^2=a^2b^2 \\ \Rightarrow \frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \\ \therefore \frac{2x}{a^2}-\frac{2y}{b^2}y'=0~~[\text{where}\,\,y'=\frac{dy}{dx}]\\ \Rightarrow \frac{x}{a^2}=\frac{yy'}{b^2} \\ \Rightarrow \frac{x}{yy'}=\frac{a^2}{b^2}\\ \therefore \frac{yy'-x[(y')^2+yy'']}{(yy')^2}=0 \\ \Rightarrow yy'-x[(y')^2+yy'']=0~~[~~\text{where}\,\,y''=\frac{d^2y}{dx^2}]\\ \Rightarrow x\left(yy''+(y')^2\right)-yy'=0\,\text{(ans.)}$

$\,20.(i)\,$ Determine the differential equation of the family parabolas whose areas are coincident with the axis of $\,x$.

Sol. The differential equation of the family parabolas whose areas are coincident with the axis of $\,x\,$ is :

$~~~~y^2=4a(x-\alpha)\\ \therefore 2yy'=4a ~~~[\text{where}\,\,y'=\frac{dy}{dx}]\\ \Rightarrow yy'=2a \\ \therefore (y')^2+yy''=0~~~[\text{where}\,\,y''=\frac{d^2y}{dx^2}]$

$\,20(ii)\,$ Form the differential equation of family of parabolas having vertex at the origin and axis along positive $\,y\,$-axis. [NCERT]

Sol. The equation of family of parabolas having vertex at the origin and axis along positive $\,y\,$-axis  is :

$~~~~~x^2=4ay \rightarrow(1) \\ \therefore \text{Differentiating (1) w.r.t. } x, \\ ~~~~2x=4ay'~~[\text{where}\,\,y'=\frac{dy}{dx}] \\ \Rightarrow x=2ay' \\ \Rightarrow 2a=\frac{x}{y'}\rightarrow(2)$

Hence, from $\,(1)\,$ and $\,(2),\,$ we get,

$~~~x^2=2 \times 2a \times y \\ \Rightarrow x^2=2 \times \frac{x}{y'} \times y\\ \Rightarrow x=\frac{2y}{y'}\\ \therefore xy'=2y~~\text{(ans.)}$

$\,21.\,$ Form the differential equation of the family of circles $~~(x-a)^2 + (y-a)^2=a^2,$ where $\,a\,$ is an arbitrary constant.

Sol. $~~(x-a)^2 + (y-a)^2=a^2 \rightarrow(1)\\ \therefore 2(x-a)+2(y-a)y'=0~~[\text{where}~~y'=\frac{dy}{dx}] \\ \Rightarrow (x-a)+(y-a)y'=0\\ \Rightarrow a+ay'=x+yy' \\ \Rightarrow a(1+y')=x+yy' \\ \Rightarrow a=\frac{x+yy'}{1+y'} \rightarrow(2)$

Now, from $\,(1)~~ \text{and}~~(2)\,\,\,$we get,

$\left(x-\frac{x+yy'}{1+y'}\right)^2+\left(y-\frac{x+yy'}{1+y'}\right)^2=\left(\frac{x+yy'}{1+y'}\right)^2 \\ \Rightarrow \left(\frac{x+xy'-x-yy'}{1+y'}\right)^2+\left(\frac{y+yy'-x-yy'}{1+y'}\right)^2\\~~~~~~~~~~=\left(\frac{x+yy'}{1+y'}\right)^2 \\ \Rightarrow (xy'-yy')^2+(y-x)^2=(x+yy')^2 \\ \Rightarrow  y'^2(y-x)^2+(y-x)^2=(x+yy')^2 \\ \Rightarrow  (y-x)^2 (1+y'^2)=(x+yy')^2~~\text{(ans.)}$