Differential Equations

Given, $se{c}^{2}xtanydx+se{c}^{2}ytanxdy=0$

Dividing throughout by ‘ $tanxtany$ ’ we get,

$\begin{array}{l}\frac{se{c}^{2}xtany}{tanxtany}dx+\frac{se{c}^{2}ytanx}{tanxtany}dy=0\\ \Rightarrow \frac{se{c}^{2}x}{tanx}dx+\frac{se{c}^{2}y}{tany}dy=0\end{array}$

Integrating both sides we get,

$\begin{array}{l}{\displaystyle \int \frac{se{c}^{2}x}{tanx}dx+{\displaystyle \int \frac{se{c}^{2}y}{tany}dy=logc}}\\ \Rightarrow log\left|tanx\right|+log\left|tany\right|=logc\left\{{\displaystyle \int \frac{f^{|}\left(x\right)}{f\left(x\right)}}dx-log\left|f\left(x\right)\right|\right\}\\ \Rightarrow log\left|\left(tanx+tany\right)\right|=logc\end{array}$

$\Rightarrow tanxtany=\pm c$ is the required general solution.

Given,  $\frac{dy}{dx}+y=1$

$\Rightarrow \frac{dy}{dx}=1-y=-\left(y-1\right)$

By separable of variable,

$\frac{dy}{\left(y-1\right)}=-dx$

Integrating both sides,

$\begin{array}{l}{\displaystyle \int \frac{dy}{\left(y-1\right)}=-{\displaystyle \int dx}}\\ \Rightarrow log\left|y-1\right|=-x+c\\ \Rightarrow \left|y-1\right|=e^{-x+c}\\ \Rightarrow y-1=\pm e^{-x}.e^c\end{array}$

$\Rightarrow y=1+Ac$ where $A=\pm e^c$

Is the general solution.

Kindly go through the solution

The given D.E is

$\frac{dy}{dx}=\frac{1-cosx}{1+cosx}$

By separable of variable,

$\begin{array}{l}\Rightarrow dy=\frac{1-cosx}{1+cosx}dx\left\{\begin{array}{l}cos2x=1-2si{n}^{2}x\\ =2si{n}^{2}x=1-cos2x\\ =2si{n}^2\frac{x}{2}=1-cosx\\ cos2x=2co{s}^{2}x-1\end{array}\right\}\\ \Rightarrow dy=\frac{2si{n}^2\frac{x}{2}}{2co{s}^2\frac{x}{2}}dx\\ \Rightarrow dy=ta{n}^2\frac{x}{2}dx\end{array}$

Integrating both sides,

$\begin{array}{l}{\displaystyle \int dy={\displaystyle \int ta{n}^2\frac{x}{2}dx\left\{se{c}^{2}x=1+ta{n}^2\right\}}}\\ \Rightarrow y={\displaystyle \int \left(se{c}^2\frac{x}{2}-1\right)}dx\end{array}$

$\Rightarrow y=\frac{tan\frac{x}{2}}{\frac{1}{2}}-x+c$ c = constant

$\Rightarrow y=2tan\frac{x}{2}-x+c$ is the general solution.

The highest order derivation present in the D.E. is y, so its order is 1.

As the given D.E. is a polynomial equation in its derivative its degree is 1.

The given equation of curve is  $y = x$ .

Differentiating with respect to x, we get:

$\frac{dy}{dx}=1 ..........(1)$

Again, differentiating with respect to x, we get:

$\frac{d^{2}y}{d{x}^2}=0..........(2)$

Now, on substituting the values of y,  $\frac{d^{2}y}{d{x}^2}$ and $\frac{dy}{dx}$   from equation (1) and (2) in each of the given alternatives, we find that

Given: $y=c_1{e}^x+c_2{e}^{-x}.........(1)$

Differentiating with respect to x, we get:

$\frac{dy}{dx}=c_1{e}^x-c_2{e}^{-x}$

Again, differentiating with respect to x, we get:

$\begin{array}{l}\frac{d^{2}y}{d{x}^2}=c_1{e}^x+c_2{e}^{-x}\\ \Rightarrow \frac{d^{2}y}{d{x}^2}=y\\ \Rightarrow \frac{d^{2}y}{d{x}^2}-y=0\end{array}$

This is the required differential equation of the given equation of curve.

Hence, the correct answer is B.

Let the centre of the circle on y-axis be (0, b).

The differential equation of the family of circles with centre at (0, b) and radius 3 is as follows:

$\begin{array}{l}{x}^2+{\left(y-b\right)}^2=3^2\\ \Rightarrow x^2+{\left(y-b\right)}^2=9..........(1)\end{array}$

Differentiating equation (1) with respect to x, we get:

$\begin{array}{l}2x+2(y-b).y^{\text{'}}=0\\ \Rightarrow (y-b).y^{\text{'}}=-x\\ \Rightarrow y-b=\frac{-x}{y^{\text{'}}}\end{array}$

Substituting the value of $(y-b)$ in equation (1), we get:

$\begin{array}{l}{x}^2+{\left(\frac{-x}{y^{\text{'}}}\right)}^2=9\\ \Rightarrow x^2\left[1+\frac{1}{{\left(y^{\text{'}}\right)}^2}\right]=9\\ \Rightarrow x^2\left({\left(y^{\text{'}}\right)}^2+1\right)=9{\left(y^{\text{'}}\right)}^2\\ \Rightarrow (x^2-9){\left(y^{\text{'}}\right)}^2+x^2=0\end{array}$

This is the required

The equation of the family of hyperbolas with the centre at origin and foci along the x-axis is:

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1..........(1)$

Differentiating both sides of equation (1) with respect to x, we get:

$\begin{array}{l}\frac{2x}{a^2}-\frac{2y{y}^{\text{'}}}{b^2}=0\\ \Rightarrow \frac{x}{a^2}-\frac{y{y}^{\text{'}}}{b^2}=0..........(2)\end{array}$

Again, differentiating both sides with respect to x, we get:

$\begin{array}{l}\frac{1}{a^2}-\frac{y^{\text{'}}.y^{\text{'}}+y.y^{"}}{b^2}=0\\ \Rightarrow \frac{1}{a^2}+\frac{1}{b^2}\left({\left(y^{\text{'}}\right)}^2+y{y}^{"}\right)=0\end{array}$

Substituting the value of $\frac{1}{a^2}$ in equation (2), we get:

$\begin{array}{l}\frac{x}{b^2}\left({\left(y^{\text{'}}\right)}^2+y{y}^{"}\right)-\frac{y{y}^{\text{'}}}{b^2}=0\\ \Rightarrow x{\left(y^{\text{'}}\right)}^2+xy{y}^{"}-y{y}^{\text{'}}=0\\ \Rightarrow xy{y}^{"}+x{\left(y^{\text{'}}\right)}^2-y{y}^{\text{'}}=0\end{array}$

This is t

The equation of the family of ellipses having foci on the y-axis and the centre at origin is as follows:

$\frac{x^2}{b^2}+\frac{y^2}{a^2}=1..........(1)$

Differentiating equation (1) with respect to x, we get:

$\begin{array}{l}\frac{2x}{b^2}+\frac{2y{y}^{\text{'}}}{b^2}=0\\ \Rightarrow \frac{x}{b^2}+\frac{y{y}^{\text{'}}}{a^2}..........(2)\end{array}$

Again, differentiating with respect to x, we get:

$\begin{array}{l}\Rightarrow \frac{1}{b^2}+\frac{y^{\text{'}}.y^{\text{'}}+y.y^{"}}{a^2}=0\\ \Rightarrow \frac{1}{b^2}+\frac{1}{a^2}\left({y^{\text{'}}}^2+y{y}^{"}\right)=0\\ \Rightarrow \frac{1}{b^2}=-\frac{1}{a^2}\left({y^{\text{'}}}^2+y{y}^{"}\right)\end{array}$

Substituting this value in equation (2), we get:

$\begin{array}{l}x\left[-\frac{1}{a^2}\left({\left(y^{\text{'}}\right)}^2+y{y}^{"}\right)\right]+\frac{y{y}^{"}}{a^2}=0\\ \Rightarrow -x{\left(y^{\text{'}}\right)}^2-xy{y}^{"}+y{y}^{\text{'}}=0\\ \Rightarrow xy{y}^{"}+x{\left(y^{\text{'}}\right)}^2=0\end{array}$

This is the required differential equat