{"id":5729,"date":"2021-09-29T18:39:37","date_gmt":"2021-09-29T13:09:37","guid":{"rendered":"https:\/\/mathemerize.com\/?p=5729"},"modified":"2021-11-21T20:11:00","modified_gmt":"2021-11-21T14:41:00","slug":"differential-equations-reducible-to-variable-separable-form","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/differential-equations-reducible-to-variable-separable-form\/","title":{"rendered":"Differential Equations Reducible to Variable Separable Form"},"content":{"rendered":"
Here you will learn how to find the solution of the differential equations reducible to variable separable form with examples.<\/p>\n
Let’s begin –<\/p>\n
Differential Equations of the form \\(dy\\over dx\\) = f(ax + by + c) can be reduce to variable separable form by the substitution ax + by + c = 0 which can be cleared by the examples given below.<\/p>\n
Example<\/strong><\/span> : Solve the differential equation : \\(sin^{-1}\\) \\(dy\\over dx\\) = x + y<\/p>\n Solution<\/span><\/strong> : We are given that,<\/p>\n \\(sin^{-1}\\) \\(dy\\over dx\\) = x + y<\/p>\n \\(\\implies\\) \\(dy\\over dx\\) = sin(x + y)<\/p>\n Let x + y = v. Then<\/p>\n 1 + \\(dy\\over dx\\) = \\(dv\\over dx\\)<\/p>\n \\(\\implies\\) \\(dy\\over dx\\) = \\(dv\\over dx\\) – 1<\/p>\n Putting x + y = v and \\(dy\\over dx\\) = \\(dv\\over dx\\) – 1 in the given differential equation, we get<\/p>\n \\(dv\\over dx\\) – 1 = sin v<\/p>\n \\(dv\\over dx\\) = 1 + sin v<\/p>\n \\(1\\over 1 + sin v\\) dv = dx<\/p>\n Now, Integrating on both sides, <\/p>\n \\(\\int\\) dx = \\(\\int\\) \\(1\\over 1 + sin v\\) dv <\/p>\n \\(\\int\\) dx = \\(\\int\\) \\(1 – sin v\\over 1 – sin^2 v\\) dv <\/p>\n \\(\\int\\) dx = \\(\\int\\) \\(1 – sin v\\over cos^2 v\\) dv <\/p>\n \\(\\int\\) dx = \\(\\int\\) \\((sec^2 v – tan v sec v)\\) dv<\/p>\n x = tan v – sec v + C<\/p>\n \\(\\implies\\) x = tan (x + y) – sec (x + y) + C, which is the required solution.<\/p>\n Example<\/strong><\/span> : Solve the differential equation : \\(dy\\over dx\\) = cos(x + y)<\/p>\n Solution<\/span><\/strong> : We are given that,<\/p>\n \\(dy\\over dx\\) = cos(x + y)<\/p>\n Let x + y = v. Then<\/p>\n 1 + \\(dy\\over dx\\) = \\(dv\\over dx\\)<\/p>\n \\(\\implies\\) \\(dy\\over dx\\) = \\(dv\\over dx\\) – 1<\/p>\n Putting x + y = v and \\(dy\\over dx\\) = \\(dv\\over dx\\) – 1 in the given differential equation, we get<\/p>\n \\(dv\\over dx\\) – 1 = cos v<\/p>\n \\(dv\\over dx\\) = 1 + cos v<\/p>\n \\(1\\over 1 + cos v\\) dv = dx<\/p>\n Now, Integrating on both sides, <\/p>\n \\(\\int\\) dx = \\(\\int\\) \\(1\\over 1 + cos v\\) dv <\/p>\n \\(\\int\\) dx = \\(\\int\\) \\(1\\over 2\\) \\(sec^2 v\/2\\) dv <\/p>\n x = \\(tan v\/2\\) + C<\/p>\n \\(\\implies\\) x = x = \\(tan (x + y)\/2\\) + C, which is the required solution<\/p>\n\n\n