Maths

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 the required differential equation.

97. Let $y=xcosx$

So,  $\frac{dy}{dx}=x\frac{d}{dx}cosx+\frac{dx}{dx}cosx$

$=-xsinx+cosx$

$\therefore \frac{d^{2}y}{d{x}^2}=-x\frac{d}{dx}sinx-sinx\frac{dx}{dx}+\frac{d}{dx}cosx$

$=-xcosx-sinx+\left(-sinx\right)$

$=-\left(xcosx+2sinx\right)$

Kindly check the Answer:

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 equation

The equation of the parabola having the vertex at origin and the axis along the positive y-axis is:

$x^2 =4ay$

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

$2x=4ay\text{'}$

Dividing equation (2) by equation (1), we get:

$\begin{array}{l}\frac{2x}{x^2}=\frac{4a{y}^{\text{'}}}{4ay}\\ \Rightarrow \frac{2}{x}=\frac{y^{\text{'}}}{y}\\ \Rightarrow x{y}^{\text{'}}=2y\\ \Rightarrow x{y}^{\text{'}}-2y=0\end{array}$

This is the required differential equation.

96. Let $y=x^{20}$

So,  $\frac{dy}{dx}=20x^{20-1}=20x^{19}$

$\therefore \frac{d^{2}y}{d{x}^2}=20*19x^{19-1}$

$=380x^{18}$

The centre of the circle touching the y-axis at origin lies on the x-axis.

Let (a, 0) be the centre of the circle.

Since it touches the y-axis at origin, its radius is a.

Now, the equation of the circle with centre (a, 0) and radius (a) is

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

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

$\begin{array}{l}2x+2y{y}^{\text{'}}=2a\\ \Rightarrow x+y{y}^{\text{'}}=a\end{array}$

Now, on substituting the value of a in equation (1), we get:

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

This is the required differential equation.

Given: Equation of the family of curves $y=e^x\left(acosx+bsinx\right)..........(i)$

Differentiating both sides with respect to x, we get:

$\begin{array}{l}{y}^{\text{'}}=e^x\left(acosx+bsinx\right)+e^x\left(-acosx+bsinx\right)\\ \Rightarrow y^{\text{'}}=e^x\left[\left(a+b\right)cosx-\left(a-b\right)sinx\right]..........(2)\end{array}$

Again, differentiating with respect to x, we get:

$\begin{array}{l}{y}^{"}=e^x\left[\left(a+b\right)cosx-\left(a-b\right)sinx\right]+e^x\left[-\left(a+b\right)sinx-\left(a-b\right)cosx\right]\\ y^{"}=e^x\left[2bcosx-2asinx\right]\\ y^{"}=2e^x\left(bcosx-asinx\right)\\ \Rightarrow \frac{y^{"}}{2}=e^x\left(bcosx-asinx\right)..........(3)\end{array}$

Adding equations (1) and (3), we get:

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

This is the required differential equation of the given curve.

95. Let y = x2 + 3x + 2

So,  $\frac{dy}{dx}=2x+3+0$  (differentiation w r t 'x')

$?\frac{d^{2}y}{d{x}^2}=2+0$  (Again “ “ ) = 2