Differential Equations

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 only the differential equation given in alternative C is correct

$\begin{array}{l}\frac{d^{2}y}{d{x}^2}-x^2\frac{dy}{dx}+xy=0-x^2.1+x.x\\ =-x^2+x^2\\ =0\end{array}$

Therefore, option (C) is correct.

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 differential equation.

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.

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.

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.

Given: Equation of the family of curves  $y=e^{2x}\left(a+bx\right)..........(1)$

Differentiating both sides with respect to x, we get:

$\begin{array}{l}{y}^{\text{'}}=2e^{2x}\left(a+bx\right)+e^{2x}.b\\ \Rightarrow y^{\text{'}}=e^{2x}\left(2a+2bx+b\right)..........(2)\end{array}$

Multiplying equation (1) with (2) and then subtracting it from equation (2), we get:

$\begin{array}{l}{y}^{\text{'}}-2y=e^{2x}\left(2a+2bx+b\right)-e^{2x}\left(2a+2bx\right)\\ \Rightarrow y^{\text{'}}-2y=b{e}^{2x}..........(3)\end{array}$

Differentiating both sides with respect to x, we get:

$y^{"}-2y^{\text{'}}=2b{e}^{2x}..........(4)$

Dividing equation (4) by equation (3), we get:

$\begin{array}{l}\frac{y^{"}-2y^{\text{'}}}{y^{\text{'}}-2y}=2\\ \Rightarrow y^{"}-2y^{\text{'}}=2y^{\text{'}}-4y\\ \Rightarrow y^{"}-4y^{\text{'}}+4y=0\end{array}$

This is the required differential equation of the given curve.