Lecture 3 [MATH 129]

 Differential Equation

Formation of Differential Equation

[Continues]

    Question: Form the differential equation of the curve represented by $y^2 - 2ay + x^2 = a^2$. where a is an arbitrary constant.

    Solution: The given equation is,

  $$y^2 - 2ay + x^2 = a^2 \tag{1.1}\label{eq:1.1}$$  

Differentiating both sides of (1.1) with respect to $x$ we get,

  $$\begin{align*} & 2y \frac{dy}{dx} - 2 a \frac{dy}{dx} + 2x = 0 \\ \implies & y \frac{dy}{dx}+ x = a \frac{dy}{dx} \\ \implies & y + x \frac{dx}{dy} = a \end{align*}$$  

Substituting this value of $a$ in (1.1) we have, 

  $$\begin{align*} & y^2 - 2h \left(y + x \frac{dx}{dy}\right) + x^2 = \left( y + x \frac{dx}{dy}\right)^2 \\ \implies & y^2 - 2y^2 - 2 xy \frac{dx}{dy} + x^2 = y^2 + 2xy \frac{dx}{dy}+ x^2 \left(\frac{dx}{dy}\right)^2 \\ \implies & 2 y^2 + 4xy \frac{dx}{dy} + x^2 \left(\frac{dx}{dy}\right)^2 - x^2 = 0 \\ \implies & 2 y^2 \left( \frac{dy}{dx}\right)^2 + 4xy \frac{dy}{dx} + x^2 - x^2 \left(\frac{dy}{dx}\right)^2 = 0 \end{align*}$$  

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    Question: Form the differential equation of all straight lines passing through the origin.

    Solution: We know, the equation of all straight lines passing through the origin is,

  $$y = mx \tag{2.1}\label{eq:2.1}$$  

Differentiating (2.1).

  $$\frac{dy}{dx} = m \tag{2.2}\label{eq:2.2}$$  

eliminating m between (2.1) and (2.2)

i.e by putting    $m = \frac{dy}{dx}$   in  $y = mx$

we get,  $$y = x \frac{dy}{dx}$$  

Which is the required differential equation. 

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    Question: Form the differential equation corresponding to \(y = ae^{2x} + be^{-3x}+ ce^x \), where $a,b,c$ are arbitrary constantx.

    Solution: The given equation is, 

&nbsp $$y = a e^{2x}+ be^{-3x} + ce^x \tag{3.1}\label{eq:3.1}$$  

Differentiating both sides of (3.1) with respect o x, we get 

  $$\frac{dy}{dx} = 2ae^{2x} - 3be^{-3x} + ce^x \tag{3.2}\label{eq:3.2}$$  

Subtracting (3.1) from (3.2) we get, 

  $$\frac{dy}{dx} - y = a e^{2x} - 4be^{-3x}$$  

Differentiating (3.3) we get,

  $$\frac{d^2 y}{d x^2} - \frac{dy}{dx} = 3ae^{2x} + 12 be^{-3x}$$  

Adding 3 times of (3.3) to (3.4) we have,

  $$\begin{align*} & \left( \frac{d^2 y}{d x^2} - \frac{dy}{dx}\right) + d \left( \frac{dy}{dx} - y\right) = (2ae^{2x}+ 12be^{-3x}) + 3 (ae^{2x} - 4be^{-3x})\\ & \frac{d^2 y}{d x^2} + 2 \frac{dy}{dx} - 3y = 5ae^{2x} \tag{3.5}\label{eq:3.5} \end{align*}$$  

Differentiating both sides of (3.5) with respect to x we get,

  $$\frac{d^3 y}{d x^3} + 2 \frac{d^2 y}{d x^2} - \frac{dy}{dx} = 10 ae^{2x} \tag{3.6}\label{eq:3.6}$$  

From (3.5) and (3.6) eliminating $a$ we get,

&nbsp $$\begin{align*} & \frac{d^3 y}{d x^3} + 2 \frac{d^2 y}{d x^2} - 3 \frac{dy}{dx} = 2 \left( \frac{d^2 y}{d x^2} + 2 \frac{dy}{dx} - 3y\right) \\ \implies & ~~ \frac{d^3 y}{d x^3} - 7 \frac{dy}{dx} + 6y = 0 \end{align*}$$  

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    Question: Find the differential equation of the family of curves $xy = Ae^x +  Be^{-x} + x^2$ .

    Solution: Given equation is, 

  $$xy = A e^x + B e^{-x} + x^2$$

differentiating with respect to x we get,

  $$x \frac{dy}{dx} + y = Ae^x - Be^{-x} + 2x$$  

Differentiating again we get,

  $$\begin{align*} & x \frac{d^2 y }{d x^2} + \frac{dy}{dx} + \frac{dy}{dx} = Ae^x + B e^{-x} + 2\\ & x \frac{d^2 y}{d x^2} + 2 \frac{dy}{dx} = xy - x^2 + 2 \end{align*}$$  

Thus the two variables A and B have been eliminated and we get the differential equation of the 2nd order as,

  $$x \frac{d^2 y}{d x^2} + 2 \frac{dy}{dx} = xy - x^2 + 2$$  

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    Question: Find the differential equation of the family of curves $y = e^x (A\cos x + B \sin x)$, where $A$ and $B$ are arbitrary constants.

    Solution: The given equation is,

  $$y = e^x (A \cos x + B \sin x) \tag{4.1}\label{eq:4.1}$$  

Differentiating with respect to x, we get

  $$\begin{align*} & \frac{dy}{dx} = e^x ( - A \sin x + B \cos x) + e^x (A \cos x + B \sin x)\\ & \frac{dy}{dx} = e^x ( - A \sin x + B \cos x) + y \tag{4.2}\label{eq:4.2} \end{align*}$$  

Again Differentiating both sides with respect to x, we get

  $$\begin{align*} \frac{d^2 y}{d x^2} & = e^x ( - A \cos x - B \sin x) + e^x ( - A\sin x + B\cos x) + \frac{dy}{dx}\\ & = - e^x ( A \cos x + B \sin x ) + \left( \frac{dy}{dx} - y\right) + \frac{dy}{dx} \\ & - y + \left( \frac{dy}{dx} - y\right) + \frac{dy}{dx} \\ \therefore~ \frac{d^2 y}{d x^2} & - 2 \frac{dy}{dx} + 2y = 0 \end{align*}$$  

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    Question: Find the differential equation of the family $y = cx^3$.

    Solution: The given equation is,

  $$\begin{align*} & y = c x^3 \\ & c = \frac{y}{x^3} \end{align*}$$  

Now 

  $$\begin{align*} & y = cx^3 \tag{5.1}\label{eq:5.1}\\ & \frac{dy}{dx} = 3cx^2 \\ & \frac{dy}{dx} = 3 \frac{y}{x^3 x^2}\\ & \frac{dy}{dx} = 3 \frac{y}{x}\\ & x \frac{dy}{dx} = 3y \\ & x \frac{dy}{dx} - 3y = 0 \\ \end{align*}$$  

which is the required differential equation.

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    Homeworks:

1. Show that the differential equation of the family of the circle touches the x-axis at the origin is  $(x^2 - y^2) ~dy - 2xy~dx = 0$  

2. Find the Differential Equation of the family of curves  $y = e^x (A\cos x + B \sin x)$ where $A$ and $B$ are arbitrary constants.

3. Discuss the degree and order of the differential equations with examples.

4. Find the differential equation of all circles of radius $a$.

5. Form a differential equation from  $y = A \cos ax + B \sin ax$ , where $A$ and $B$ arbitrary constants and $a$ is a fixed constant.

6. Find the differential equation of all circles which pass through the origin and whose centers are on the x-axis.

7. briefly discuss differential equation with its degree and order. Form an ordinary differential equation from  $y = e^x(a \cos x + b \sin x)$ , where $a$ and $b$ parameter.