Differentiation (Part-38) | S N De

 

Form the differential equation by eliminating $\,A\,$ and $\,B\,$ (or $\,a\,$ or $\,b\,$) [$\,10-12\,$]: 

$\,10.~~(y-b)^2=4k(x-a)$

Sol. $~~(y-b)^2=4k(x-a)\\ \therefore 2(y-b)y'=4k~~~[\text{where}~~~y'=\frac{dy}{dx}] \\ \Rightarrow (y-b)y'=2k \rightarrow(1) \\ \Rightarrow (y')^2+(y-b)y''=0~~[\text{where}~~y''=\frac{d^2y}{dx^2}] \\~~~~~~~~~~~\rightarrow(2)$

So, from $\,(1),(2)~~$ we get,

$\,(y')^2+\frac{2k}{y'}.y''=0 \\ \therefore 2ky''+(y')^3=0~~~\text{(ans.)}$

$\,11.~~y=a\sec x+b\tan x$

Sol. $~~~~~y=a\sec x+b\tan x\rightarrow(1) \\ \text{Differentiating (1) w.r.t }~~x, \\ \Rightarrow y'=a\sec x\tan x+b\sec^2x \\~~~~~~~~~~~[\text{where}~~~y'=\frac{dy}{dx}]\\ \therefore y'\cos x=a\tan x+b\sec x  \rightarrow(2)\\ \text{Differentiating (2) w.r.t }~~x,\\ (-\sin x) y'+(\cos x)y''=a\sec^2x\\+b\sec x\tan x~~~~~~~~[\text{where}~~y''=\frac{d^2y}{dx^2}]\\ \therefore (-\sin x\cos x) y'+(\cos^2 x)y'' \\~~~~=a\sec x+b\tan x \\ \Rightarrow (-\sin x \cos x)y'+(\cos^2x)y''=y~~~[\text{By (1)}] \\ \Rightarrow (-\tan x)y'+y''=y\sec^2x \\ \therefore y''=(\tan x)y'+y\sec^2x~~\text{(ans.)}$

$\,12.~~~xy=ae^x+Be^{-x}$

Sol. $~~~xy=Ae^x+Be^{-x}\rightarrow(1) \\ \therefore y+xy'=Ae^x-Be^{-x}\\~~~~ \therefore y'+y'+xy''=Ae^x+Be^{-x}\\ \Rightarrow xy''+2y'=xy~~[\text{By (1)}]~~\text{(ans.)}$

$\,13.~~$ Show that the differential equation $\,x[yy_2+(y_1)^2]=yy_1~~~$ is formed by eliminating $\,a,b\,$ and $\,c\,$ from the relation $\,\,ax^2+by^2+c=0.$ Justify why the eliminate is of the second order although the given relation involves three constants $\,a,b\,$ and $\,c.$

Sol. Here, $\,y_1=y'=\frac{dy}{dx},~~y_2=y''=\frac{d^2y}{dx^2}.$

We have , $\,ax^2+by^2+c=0 \\ \Rightarrow by^2=-ax^2-c \\ \Rightarrow y^2=-\frac ab x^2-\frac cb \rightarrow(1)\\ \therefore \text{Differentiating (1) w.r.t }~~x, \\ ~~~~2yy'=-\frac ab \times 2x \\ \therefore yy'=-\frac abx\rightarrow(2)\\ \therefore \text{Again differentiating (2) w.r.t }~~x,\\ ~~~(y')^2+yy''=-\frac ab\rightarrow(3) $

Now, from $\,(2),(3)~~$, we get,

$~~~~yy'=[(y')^2+yy'']x\\ \Rightarrow x(yy_2+y_1^2)=yy_1\rightarrow(4)$

We notice that $\,(1)\,$ can be written as $\,y^2=k_1x^2+k_2,\,$ where $\,k_1=-\frac ab, k_2=-\frac cb.$

Again, we know that the order of the differential equation is the order of the highest derivative (or differential) present in the equation.

So, from $\,(4),\,$ it follows that the differential equation is of $\,2$nd order.

$\,14.~~$ Show that, the solution $\,x=A\cos(nt+B)+\frac{k}{n^2-p^2}.\sin pt,\,\,$ for all $\,A\,$ and $\,B\,$ satisfies the differential equation $\,\,\frac{d^2x}{dt^2}+n^2x=k\sin pt.$

Sol. $~~~~x=A\cos(nt+B)+\frac{k}{n^2-p^2}.\sin pt\\ \therefore \frac{dx}{dt}=-An\sin(nt+B)+\frac{kp}{n^2-p^2}\cos pt \\ \therefore \frac{d^2x}{dt^2}=-n^2A\cos(nt+B)-\frac{kp^2}{n^2-p^2}\sin pt$

Now, $\,\frac{d^2x}{dt^2}+n^2x\\=-n^2A\cos(nt+B)-\frac{kp^2}{n^2-p^2}\sin pt \\~~+n^2A\cos(nt+B)+\frac{n^2k}{n^2-p^2}\sin pt\\=\frac{k(n^2-p^2)}{n^2-p^2}\sin pt\\=k\sin pt \rightarrow(1)$

Hence, from $\,(1),\,$ we can conclude that the solution $\,x=A\cos(nt+B)+\frac{k}{n^2-p^2}.\sin pt,\,\,$ for all $\,A\,$ and $\,B\,$ satisfies the differential equation $\,\,\frac{d^2x}{dt^2}+n^2x=k\sin pt.$

$\,15.\,$ Show that, the solution  $\,x=e^{-kt}(a\cos nt+b\sin nt),\,$ for all $\,a\,$ and $\,b\,$, always satisfies the differential equation $\,\,\frac{d^2x}{dt^2}+2k\frac{dx}{dt}+(k^2+n^2)x=0$

Sol. $~~~~~x=e^{-kt}(a\cos nt+b\sin nt) \\ \Rightarrow xe^{kt}=a\cos nt+b \sin nt \rightarrow(1)\\ \therefore e^{kt}\frac{dx}{dt}+kxe^{kt}=n(-a\sin nt+b\cos nt)\\ \therefore ke^{kt}\frac{dx}{dt}+e^{kt}\frac{d^2x}{dt^2}+ke^{kt}\frac{dx}{dt}+k^2xe^{kt}\\=-n^2(a\cos nt+b\sin nt)\\=-n^2xe^{kt}~~~[\text{By (1)}]\\ \Rightarrow e^{kt}\frac{d^2x}{dt^2}+2ke^{kt}\frac{dx}{dt}+(k^2+n^2)xe^{kt}=0\\ \Rightarrow \frac{d^2x}{dt^2}+2k\frac{dx}{dt}+(k^2+n^2)x=0 \rightarrow(2)$

Hence, from $\,(2),\,$ we can conclude that the solution  $\,x=e^{-kt}(a\cos nt+b\sin nt),\,$ for all $\,a\,$ and $\,b\,$, always satisfies the differential equation $\,\,\frac{d^2x}{dt^2}+2k\frac{dx}{dt}+(k^2+n^2)x=0$

$\,16.\,$ Show that the solution $\,\,y=a\sin x+b\cos x+x\sin x\,\,$ satisfies , $\,\frac{d^2y}{dx^2}+y=2\cos x.$

Sol. $~~~~~~y=a\sin x+b\cos x+x\sin x\\ \Rightarrow y-x\sin x=a\sin x+b\cos x\rightarrow(1)\\ \therefore \frac{dy}{dx}-\sin x-x\cos x=a\cos x-b\sin x \\ \therefore \frac{d^2y}{dx^2}-\cos x-\cos x+x\sin x\\=-(a\sin x+b\cos x)\\ \Rightarrow \frac{d^2y}{dx^2}-2\cos x+x\sin x=-(y-x\sin x)\\~~~~~~~~~~~~~~~~~~~~~~[\text{By (1)}]\\ \therefore \frac{d^2y}{dx^2}-2\cos x+x\sin x=-y+x\sin x \\ \therefore \frac{d^2y}{dx^2}+y=2\cos x \rightarrow(2)$

Hence, by $\,(2),\,$ we can conclude that  the solution $\,\,y=a\sin x+b\cos x+x\sin x\,\,$ satisfies , $\,\frac{d^2y}{dx^2}+y=2\cos x.$

$\,17.\,$ Show that, the equation of all circles touching the $\,y-$axis at the origin is, $\,2xy \frac{dy}{dx}=y^2-x^2.$

Sol. The equation of circles touching the $\,y-$axis at the origin is :

$\,(x-a)^2+y^2=a^2,\,[\,\text{a being the radius.}] \\ \Rightarrow x^2-2xa+a^2+y^2=a^2 \\ \Rightarrow x^2+y^2-2ax=0 \rightarrow(1) \\ \text{Differentiating (1) w.r.t. to }\,x, \\~~~~ 2x+2y\frac{dy}{dx}-2a=0 \\ \Rightarrow x+y\frac{dy}{dx}-a=0 \\ \Rightarrow a=x+y\frac{dy}{dx}\rightarrow(2)$

Now, from $\,(1),(2)~$, we get,

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