{"id":5727,"date":"2021-09-29T18:43:29","date_gmt":"2021-09-29T13:13:29","guid":{"rendered":"https:\/\/mathemerize.com\/?p=5727"},"modified":"2021-11-21T20:11:28","modified_gmt":"2021-11-21T14:41:28","slug":"differential-equations-in-variable-separable-form","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/differential-equations-in-variable-separable-form\/","title":{"rendered":"Differential Equations in Variable Separable Form"},"content":{"rendered":"
Here you will learn how to find the solution of the differential equations in variable separable form with examples.<\/p>\n
Let’s begin –<\/p>\n
If the differential equation can be put in the form f(x) dx = g(y) dy, we say that the variables are seperable and such equations can be solved by integrating on both sides. The solution is given by <\/p>\n
\n\\(\\int\\) f(x) dx = \\(\\int\\) g(y) dy + C, where C is an arbitrary constant<\/p>\n<\/blockquote>\n
Note<\/strong> : There is no need of introducing arbitrary constants of integration on both sides as they can be combined together to give just one arbitrary constant.<\/p>\n
Example<\/strong><\/span> : Solve the differential equation : (x + 1)\\(dy\\over dx\\) = 2xy<\/p>\n
Solution<\/span><\/strong> : We have, <\/p>\n
(x + 1)\\(dy\\over dx\\) = 2xy<\/p>\n
\\(\\implies\\) (x + 1)dy = 2xy dx<\/p>\n
\\(\\implies\\) \\(dy\\over y\\) = \\(2x\\over x + 1\\) dx<\/p>\n
Now, integrating on both sides,<\/p>\n
\\(\\int\\) \\(1\\over y\\) dy = 2 \\(\\int\\) \\(x\\over x + 1\\) dx<\/p>\n
\\(\\implies\\) \\(1\\over y\\) dy = 2 \\(\\int\\) \\(x + 1 – 1\\over x + 1\\) dx<\/p>\n
\\(\\implies\\) \\(1\\over y\\) dy = 2 \\(\\int\\) \\(1 – {1\\over x + 1}\\) dx<\/p>\n
log y = 2{x – log| x + 1 |} + C, which is the solution of the given differential equation.<\/p>\n
Example<\/strong><\/span> : Solve the differential equation : cos x(1 + cos y) dx – sin y(1 + sin x)dy = 0<\/p>\n
Solution<\/span><\/strong> : We have, <\/p>\n
cos x(1 + cos y) dx – sin y(1 + sin x)dy = 0<\/p>\n
\\(\\implies\\) \\(cos x\\over 1 + sin x\\) dx – \\(sin y\\over 1 + cos y\\) dy = 0<\/p>\n
\\(\\implies\\) \\(cos x\\over 1 + sin x\\) dx + \\(-sin y\\over 1 + cos y\\) dy = 0<\/p>\n
Now, integrating on both sides,<\/p>\n
\\(\\int\\) \\(cos x\\over 1 + sin x\\) dx + \\(\\int\\) \\(-sin y\\over 1 + cos y\\) dy = 0<\/p>\n
\\(\\implies\\) log|1 + sin x| + log|1 + cos y| = log C<\/p>\n
\\(\\implies\\) log{|1 + sin x|.|1 + cos y|} = log C<\/p>\n
|1 + sin x|.|1 + cos y| = C<\/p>\n
Hence, (1 + sin x)(1 + cos y) = C is the solution of the given differential equation.<\/p>\n\n\n