{"id":5700,"date":"2021-09-29T18:36:19","date_gmt":"2021-09-29T13:06:19","guid":{"rendered":"https:\/\/mathemerize.com\/?p=5700"},"modified":"2021-11-21T20:15:58","modified_gmt":"2021-11-21T14:45:58","slug":"formation-of-differential-equation","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/formation-of-differential-equation\/","title":{"rendered":"Formation of Differential Equation"},"content":{"rendered":"
Here you will learn formation of differential equation with examples.<\/p>\n
Let’s begin –<\/p>\n
Algorithm<\/strong><\/p>\n 1). Write the equation involving independent variable x (say), dependent variable y (say) and the arbitrary constants.<\/p>\n 2). Obtain the number of arbitrary constants in Step 1. Let there be n arbitrary constants.<\/p>\n 3). Differentiate the relation in step 1 n times with respect to x.<\/p>\n 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.<\/p>\n<\/blockquote>\n Example<\/span><\/strong> : Form the differential equation of the family of curves represented by \\(c(y + c)^2\\) = \\(x^3\\) , where c is a parameter.<\/p>\n Solution<\/span><\/strong> : The equation of the family of curves is \\(c(y + c)^2\\) = \\(x^3\\) ……….(i)<\/p>\n Clearly, it is one parameter family of curves, so we shall get a differential equation of first order.<\/p>\n Differentiating (i) with respect to x, we get<\/p>\n 2c(y + c) \\(dy\\over dx\\) = \\(3x^2\\) ………(ii)<\/p>\n Dividing (i) by (ii), we get<\/p>\n \\(c(c + y)^2\\over 2c(y + c){dy\\over dx}\\) = \\(x^3\\over 3x^2\\)<\/p>\n \\(\\implies\\) y + c = \\(2x\\over 3\\)\\(dy\\over dx\\)<\/p>\n \\(\\implies\\) c = \\(2x\\over 3\\)\\(dy\\over dx\\) – y<\/p>\n Substituting the value of c in (i), we get<\/p>\n (\\(2x\\over 3\\)\\(dy\\over dx\\) – y)\\(({2x\\over 3}{dy\\over dx})^2\\) = \\(x^3\\)<\/p>\n \\(\\implies\\) \\(8\\over 27\\)\\(x({dy\\over dx})^3\\) – \\(4\\over 9\\)\\(({dy\\over dx})^2\\)y = x<\/p>\n \\(\\implies\\) \\(8x({dy\\over dx})^3\\) – \\(12y({dy\\over dx})^2\\) = 27x<\/p>\n This is the required differential equation of the curves represented by (i).<\/p>\n Example<\/span><\/strong> : Form the differential equation representing the family of curves y = A cos(x + B), where A and B are parameter.<\/p>\n Solution<\/span><\/strong> : The equation of the family of curves is y = A cos(x + B) ……….(i)<\/p>\n This equation contains two arbitratry constants. So, let us differential it two times to obtain a differential equation of second order.<\/p>\n Differentiating (i) with respect to x, we get<\/p>\n \\(dy\\over dx\\) = -A sin(x + B) ………(ii)<\/p>\n Differentiating (ii) with respect to x, we get<\/p>\n \\(d^2y\\over dx^2\\) = -A cos(x + B)<\/p>\n \\(\\implies\\) \\(d^2y\\over dx^2\\) = -y [Using (i)]<\/p>\n \\(\\implies\\) \\(d^2y\\over dx^2\\) + y = 0, which is the required differential equation of the given family of curves. <\/p>\n\n\n\n