Differentiation (Part-38) | S N De

 $\,22.\,$ Show that the function $\,y=A\cos2x-B\sin2x\,$ is a solution of the differential equation $~~\frac{d^2y}{dx^2}+4y=0.~$ [CBSE-'07]

Sol. $~~~~~y=A\cos2x-B\sin2x\\ \therefore y'=2(-A\sin2x-B\cos2x)~~~[y=\frac{dy}{dx}]\\ \therefore y''=-4(A\cos2x-B\sin2x)~~~[y''=\frac{d^2y}{dx^2}] \\ \Rightarrow y''=-4y \\ \Rightarrow y''+4y=0\rightarrow(1)$

Hence, from $\,(1),\,$ we can conclude that  the function $\,y=A\cos2x-B\sin2x\,$ is a solution of the differential equation $~~\frac{d^2y}{dx^2}+4y=0.$

$\,23.\,$  Form the differential equation of the family of circles having centre on the $\,x\,$-axis and passing through the origin. 

Sol. Let the center of the circle be  $\,(a,0)\,$ and $\,r\,$ be the radius of the circle.

So, the equation of the circle is given by :

$~~~(x-a)^2+y^2=r^2 \\ \Rightarrow x^2+y^2-2ax+a^2-r^2=0\rightarrow(1)$

Since the circle passes through the point $\,(0,0),$ we get from $\,(1),$

$\,a^2-r^2=0\rightarrow(2)$

So, from $\,(1),(2)~$, we get 

$~~~~x^2+y^2-2ax=0 \rightarrow(3)\\ \text{So, differentiating  (3) w.r.t}~~x, \\ ~~~2x+2yy'-2a=0~~[\text{where}~~y'=\frac{dy}{dx}]\\ \Rightarrow x+yy'-a=0\\ \Rightarrow a=x+yy' \rightarrow(4)$

Hence, from $\,(3),(4)~~,$we get,

$~~x^2+y^2-2x(x+yy')=0 \\ \Rightarrow x^2+y^2-2x^2-2xyy'=0\\ \Rightarrow y^2-x^2=2xyy'~~\text{(ans.)}$

$\,24.\,$ Form the differential equation representing the family of ellipses having centre at the origin and foci on $\,y\,$-axis.

Sol. The equation representing the family of ellipses having centre at the origin and foci on $\,y\,$-axis is given by :

$~~~~~\frac{x^2}{b^2}+\frac{y^2}{a^2}=1 \rightarrow (1)\\ \therefore \text{differentiating (1) w.r.t.} ~~ x, \\ ~~~\frac{2x}{b^2}+\frac{2yy'}{a^2}=0~~[\text{where}~~y'=\frac{dy}{dx}]\\ \Rightarrow \frac{x}{b^2}+\frac{yy'}{a^2}=0 \\ \Rightarrow \frac{yy'}{x}=-\frac{a^2}{b^2}\rightarrow(2)\\ \text{Again, differentiating (2) w.r.t. }~~x, \\~~\frac{x(yy''+(y')^2)-yy'}{x^2}=0~~[~~\text{where}~~y''=\frac{d^2y}{dx^2}]\\ \Rightarrow x(yy''+(y')^2)-yy'=0 \\ \Rightarrow x(yy''+(y')^2)=yy'\,\,\,\text{(ans.)}$

$\,25.\,$ Form the differential equation of the family of circles in the second quadrant and touching the coordinate axes. [NCERT] 

Sol. The equation of the family of circles in the second quadrant and touching the coordinate axes is given by :

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

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

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

$\,26.\,$ Form the differential equation that represents parabolas each of which has a latus rectum $\,4a\,$, and whose axes are parallel to the $\,x\,$-axis.

Sol. The equation that represents parabolas each of which has a latus rectum $\,4a\,$, and whose axes are parallel to the $\,x\,$-axis, is given by : 

$~~~(y-\beta)^2=4a(x-\alpha)$

Now follow the solution of Short Answer Type Question no. $\,10.$

$\,27.\,$ If $\,a\,$ is a parameter, show that the differential equation of the family of curves $\,\,\,y =\frac{a-x}{ax+1}\,\,$ is $\,\,(x^2+1)\frac{dy}{dx}+y^2+1=0$

Sol. Here, we denote $~~y'=\frac{dy}{dx}.$

Now, $~~~~~~y =\frac{a-x}{ax+1}\\ \Rightarrow axy+y=a-x \\ \Rightarrow a(xy-1)=-x-y\\ \Rightarrow -a(1-xy)=-(x+y)\\ \Rightarrow a=\frac{x+y}{1-xy}\rightarrow(1) \\ \therefore ~~\text{Differentiating (1) w.r.t.}~~x,\\~~~0=\frac{(1-xy)(1+y')-(x+y)(0-xy'-y)}{(1-xy)^2}\\ \Rightarrow (1-xy)(1+y')+(x+y)(y+xy')=0\\ \Rightarrow 1+y'-xy-xyy'+xy+x^2y'+y^2+xyy'=0\\ \Rightarrow 1+y'+x^2y'+y^2=0\\ \Rightarrow (1+x^2)y'+(y^2+1)=0~~~\text{(ans.)}$