Differentiation (Part-38) | S N De

  Form the differential equations by eliminating $\,a\,$ (or $\,m\,$ or $\,c\,$) [$\,\,1-6$] 

$\,1.\, y^2=4ax$

Sol. $\, y^2=4ax \rightarrow(1) \\ \therefore 2yy_1=4a~~[\text{where}~~y_1=\frac{dy}{dx}]$

Now, putting the value of $\,\,4a\,\,$ in $\,(1),\,$ we get,

$\,~~~~y^2=(2yy_1)x \\ \Rightarrow  y=2xy_1~~~[\because y \neq 0]$

$\,2.~~y=mx+5 \rightarrow(1) \\ \therefore y_1=m ~~[\text{where}~~y_1=\frac{dy}{dx}]$

Now, putting the value of $\,\,m\,\,$ in $\,(1),\,$ we get,

$\,~~~y=xy_1+5 \\ \Rightarrow xy_1-y+5=0\,\,\text{(ans.)}$

$\,3.~~x^2+y^2=c^2$

Sol. $\,~~x^2+y^2=c^2 \\ \Rightarrow 2x+2yy_1=0 \\ \Rightarrow 2\left(x+yy_1\right)=0 \\ \Rightarrow x+yy_1=0\,\,\text{(ans.)}$

$\,4.\, y=a\cos2x$

Sol. $~~y=a\cos2x \rightarrow(1)\\ \Rightarrow y_1=-2a\sin2x \\ \Rightarrow a=-\frac 12 (\csc2x )y_1$

Now, putting the value of $\,\, a\,\,$ in $\,(1),\,$ we get,

$\,y=[-\frac 12 (\csc2x)y_1] \cos2x \\ \Rightarrow y+\frac 12 \left(\frac{\cos2x}{\sin2x}\right)y_1=0 \\ \Rightarrow y+\frac 12 (\cot2x)y_1=0\\ \Rightarrow2y+(\cot2x)y_1=0 \\ \Rightarrow y_1+2y\tan2x=0\,\,\text{(ans.)}$

$\,5.\, y=c\log x-2$

Sol. $\, y=c\log x-2 \rightarrow(1) \\ \therefore y_1=\frac cx \\ \Rightarrow c=xy_1.$

Now, putting the value of $\,\,c,\,$ in $\,(1),\,$ we get,

$\,~~~~y=(xy_1)\log x-2 \\ \Rightarrow x \log x \cdot y_1=y+2\,\,\text{(ans.)}$

$\,6.\,~~y=e^{mx}$ 

Sol. $\,y=e^{mx} \rightarrow(1) \\ \therefore y_1=me^{mx}=my \\ \Rightarrow m=\frac{y_1}{y}$

$\,(1)\,$ can be rewritten as $\,\log y=mx$

Now, putting the value of $\,\,m\,\,$ in $\,(1),\,$ we get,

$\, \log y=mx=\left(\frac{y_1}{y}\right)x \\ \Rightarrow xy_1=y\log y\,\,\text{(ans.)}$

$\,7.\,$ Show that, $\,v=\frac Ar+B\,\,$ satisfies the differential equation 

$\,\,\frac{d^2v}{dr^2}+\frac 2r \cdot \frac{dv}{dr}=0.$

Sol. By question, $\,~~~~~~~v=\frac Ar+B \\ ~~\therefore \frac{dv}{dr}=-\frac{A}{r^2} \\ \Rightarrow r^2\frac{dv}{dr}=-A \\ \therefore r^2\frac{d^2v}{dr^2}+2r\frac{dv}{dr}=0 \\ \Rightarrow \frac{d^2v}{dr^2}+\frac 2r \frac{dv}{dr}=0.$ 

$\,8.\,$ Show that the differential equation $\,\,\frac{dy}{dx}=y\,\,$ is formed by eliminating $\,a\,$ and $\,b\,$ from the relation $\,y=ae^{b+x}\,$ . Justify why the eliminate is of the first order although the relation involves two constants $\,a\,$ and $\,b.$

Sol. $~~~y=ae^{b+x} \rightarrow(1) \\ \therefore \frac{dy}{dx}=ae^{b+x}=y $

We notice that $\,(1)\,$ can be written as $\,y=ae^b.e^x=Ae^x \rightarrow(2)$

From $\,(2),\,$ we observe that the equation contains only one constant and so we can say that the eliminate is of the first order although the relation involves two constants $\,a\,$ and $\,b.$

$\,9.\,$ Prove that , $\,x=A\cos \sqrt{\mu}t\,\,$ is a solution of the differential equation $\,\,\frac{d^2x}{dt^2}+\mu x=0.$

Sol. By question, $\,x=A\cos\sqrt{\mu}t \\ \therefore \frac{dx}{dt}=-A\sqrt{\mu}\sin \sqrt{\mu}t \\ \therefore \frac{d^2x}{dt^2}=-A\sqrt{\mu} \sqrt{\mu}\cos\sqrt{\mu}t\\~~~~~~~~~~=-A\mu\cos{\sqrt{\mu}t}\\~~~~~~~~~~=-\mu(A \cos\sqrt{\mu}t)\\~~~~~~~~~~=-\mu x \\ \Rightarrow \frac{d^2x}{dt^2}+\mu x=0 \rightarrow(1)$

Hence, from $\,(1),\,$ we can say that $\,x=A\cos \sqrt{\mu}t\,\,$ is a solution of the differential equation $\,\,\frac{d^2x}{dt^2}+\mu x=0.$