Formation of Differential Equation

Here you will learn formation of differential equation with examples.

Let’s begin –

Formation of Differential Equation

Algorithm

1). Write the equation involving independent variable x (say), dependent variable y (say) and the arbitrary constants.

2). Obtain the number of arbitrary constants in Step 1. Let there be n arbitrary constants.

3). Differentiate the relation in step 1 n times with respect to x.

4). Eliminate arbitrary constants with the help of n equations involving differential coefficients obtained in step 3 and an equation in step 1. The equation so obtained is the desired differential equation.

Example : Form the differential equation of the family of curves represented by \(c(y + c)^2\) = \(x^3\) , where c is a parameter.

Solution : The equation of the family of curves is \(c(y + c)^2\) = \(x^3\)                            ……….(i)

Clearly, it is one parameter family of curves, so we shall get a differential equation of first order.

Differentiating (i) with respect to x, we get

2c(y + c) \(dy\over dx\) = \(3x^2\)                          ………(ii)

Dividing (i) by (ii), we get

\(c(c + y)^2\over 2c(y + c){dy\over dx}\) = \(x^3\over 3x^2\)

\(\implies\) y + c = \(2x\over 3\)\(dy\over dx\)

\(\implies\) c = \(2x\over 3\)\(dy\over dx\) – y

Substituting the value of c in (i), we get

(\(2x\over 3\)\(dy\over dx\) – y)\(({2x\over 3}{dy\over dx})^2\) = \(x^3\)

\(\implies\) \(8\over 27\)\(x({dy\over dx})^3\) – \(4\over 9\)\(({dy\over dx})^2\)y = x

\(\implies\) \(8x({dy\over dx})^3\) – \(12y({dy\over dx})^2\) = 27x

This is the required differential equation of the curves represented by (i).

Example : Form the differential equation representing the family of curves y = A cos(x + B), where A and B are parameter.

Solution : The equation of the family of curves is y = A cos(x + B)                            ……….(i)

This equation contains two arbitratry constants. So, let us differential it two times to obtain a differential equation of second order.

Differentiating (i) with respect to x, we get

\(dy\over dx\) = -A sin(x + B)                          ………(ii)

Differentiating (ii) with respect to x, we get

\(d^2y\over dx^2\) = -A cos(x + B)

\(\implies\)  \(d^2y\over dx^2\) = -y             [Using (i)]

\(\implies\)  \(d^2y\over dx^2\) + y  = 0, which is the required differential equation of the given family of curves.                

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