{"id":5723,"date":"2021-09-29T18:38:49","date_gmt":"2021-09-29T13:08:49","guid":{"rendered":"https:\/\/mathemerize.com\/?p=5723"},"modified":"2021-11-21T20:12:14","modified_gmt":"2021-11-21T14:42:14","slug":"differential-equations-of-form-dy-dx-fx-or-fy","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/differential-equations-of-form-dy-dx-fx-or-fy\/","title":{"rendered":"Differential Equations of Form dy\/dx = f(x) or f(y)"},"content":{"rendered":"
Here you will learn how to find the general solution of differential equations of form dy\/dx = f(x) or f(y) with examples.<\/p>\n
Let’s begin –<\/p>\n
To solve this type of differential equations we integrate both sides to obtain the general solution as discussed below.<\/p>\n
\nWe have. <\/p>\n
\\(dy\\over dx\\) = f(x) \\(\\iff\\) dy = f(x)dx<\/p>\n
Integrating both sides, we obtain<\/p>\n
\\(\\int \\) dy = \\(\\int \\) f(x) dx + C or,<\/p>\n
y = \\(\\int \\) f(x) dx + C, <\/p>\n
which gives general solution of the differential equation.<\/p>\n<\/blockquote>\n
Example<\/span><\/strong> : Solve the given differential equation : \\(dy\\over dx\\) = \\(x\\over x^2 + 1\\)<\/p>\n
Solution<\/span><\/strong> : We have,<\/p>\n
\\(dy\\over dx\\) = \\(x\\over x^2 + 1\\)<\/p>\n
\\(\\implies\\) dy = \\(x\\over x^2 + 1\\)dx<\/p>\n
Integrating both sides, we get<\/p>\n
\\(\\int \\) dy = \\(\\int \\) \\(x\\over x^2 + 1\\)dx<\/p>\n
\\(\\implies\\) dy = \\(1\\over 2\\) \\(2x\\over x^2 + 1\\)dx<\/p>\n
\\(\\implies\\) y = \\(1\\over 2\\) \\(log|x^2 + 1|\\) + C<\/p>\n
Clearly, y = \\(1\\over 2\\) \\(log|x^2 + 1|\\) + C is defined for all x \\(\\in\\) R.<\/p>\n
Hence, y = \\(1\\over 2\\) \\(log|x^2 + 1|\\) + C, x \\(\\in\\) R is the solution of the given differential equation.<\/p>\n
(2) Differential Equations of Form dy\/dx = f(y)<\/strong><\/h4>\n
To solve this type of differential equations we integrate both sides to obtain the general solution as discussed below.<\/p>\n
\nWe have. <\/p>\n
\\(dy\\over dx\\) = f(y) <\/p>\n
\\(\\implies\\) \\(dx\\over dy\\) = \\(1\\over f(y)\\), provided that f(y) \\(\\ne\\) 0<\/p>\n
\\(\\implies\\) dx = \\(1\\over f(y)\\) dy<\/p>\n
Integrating both sides, we obtain<\/p>\n
\\(\\int \\) dx = \\(\\int \\) \\(1\\over f(y)\\) dy + C or,<\/p>\n
x = \\(\\int \\) \\(1\\over f(y)\\) dy + C, <\/p>\n
which gives general solution of the differential equation.<\/p>\n<\/blockquote>\n
Example<\/span><\/strong> : Solve the given differential equation : \\(dy\\over dx\\) = \\(1\\over y^2 + sin y\\)<\/p>\n
Solution<\/span><\/strong> : We have,<\/p>\n
\\(dy\\over dx\\) = \\(1\\over y^2 + sin y\\)<\/p>\n
\\(\\implies\\) \\(dx\\over dy\\) = \\(y^2 + sin y\\)<\/p>\n
\\(\\implies\\) dx = \\(y^2 + sin y\\)dx<\/p>\n
Integrating both sides, we get<\/p>\n
\\(\\int \\) dx = \\(\\int \\) \\(y^2 + sin y\\)dx<\/p>\n
\\(\\implies\\) x = \\(y^3\\over 3\\) – cosy + C<\/p>\n
Hence, x = \\(y^3\\over 3\\) – cosy + C is the solution of the given differential equation.<\/p>\n\n\n