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*}
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.
Solution: The given equation is,
  $$ y = a e^{2x}+ be^{-3x} + ce^x \tag{3.1}\label{eq:3.1} $$
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,
  \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*}
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,
$$ x \frac{d^2 y}{d x^2} + 2 \frac{dy}{dx} = xy - x^2 + 2 $$
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*}
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.
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.
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