{"id":5731,"date":"2021-09-29T18:46:31","date_gmt":"2021-09-29T13:16:31","guid":{"rendered":"https:\/\/mathemerize.com\/?p=5731"},"modified":"2021-11-21T20:07:56","modified_gmt":"2021-11-21T14:37:56","slug":"solution-of-homogeneous-differential-equation","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/solution-of-homogeneous-differential-equation\/","title":{"rendered":"Solution of Homogeneous Differential Equation"},"content":{"rendered":"
Here you will learn how to find solution of homogeneous differential equation of first order first degree with examples.<\/p>\n
Let’s begin –<\/p>\n
If a first degree first order differential equation is expressible in the form<\/p>\n
\n\\(dy\\over dx\\) = \\(f(x, y)\\over g(x, y)\\)<\/p>\n<\/blockquote>\n
where f(x, y) and g(x, y) are homogeneous function of the same degree, then it is called a homogeneous differential equation,<\/p>\n
Such type of equation can be reduced to variable seperable form by the substitution y = vx. Process is shown in the algorithm below.<\/p>\n
Algorithm :<\/strong><\/p>\n
\n1). Put the differential equation in the form<\/p>\n
\\(dy\\over dx\\) = \\(f(x, y)\\over g(x, y)\\)<\/p>\n
2). Put y = vx and \\(dy\\over dx\\) = v + x\\(dy\\over dx\\) in the equation in step 1 and cancel out x from the right hand side.<\/p>\n
3). Shift v on RHS and seperate the variables v and x.<\/p>\n
4). Integrate both sides to obtain the solution in terms of v and x.<\/p>\n
5). Replace v by \\(y\\over x\\) in the solution obtained in step 4 to obtain the solution in terms of x and y.<\/p>\n<\/blockquote>\n
Example<\/span><\/strong> : Solve the differential equation \\(x^2\\) dy + y(x + y) dx = 0.<\/p>\n
Solution<\/span><\/strong> : The given differential equation is<\/p>\n
\\(x^2\\) dy + y(x + y) dx = 0<\/p>\n
\\(\\implies\\) \\(x^2\\) dy = -y(x + y) dx<\/p>\n
\\(\\implies\\) \\(dy\\over dx\\) = -(\\(xy + y^2\\over x^2\\)) …………………(i)<\/p>\n
Since each of the functions xy + \\(y^2\\) and \\(x^2\\) is a homogeneous function of degree 2.<\/p>\n
Therefore, equation (i) is a homogeneous differential equation.<\/p>\n
Putting y = vx and \\(dy\\over dx\\) = v + x\\(dv\\over dx\\) in (i), we get<\/p>\n
v + x\\(dv\\over dx\\) = -(\\(vx^2 + v^2x^2\\over x^2\\))<\/p>\n
v + x\\(dv\\over dx\\) = -(\\(v + v^2\\))<\/p>\n
x\\(dv\\over dx\\) = -(\\(2v + v^2\\))<\/p>\n
xdv = -(\\(v^2 + 2v\\))dx<\/p>\n
By Seperating the variable,<\/p>\n
\\(1\\over v^2 + 2v\\) dv = \\(-dx\\over x\\)<\/p>\n
Integrating both sides.<\/p>\n
\\(\\implies\\) \\(\\int\\) \\(1\\over v^2 + 2v\\) dv = \\(\\int\\) \\(-1\\over x\\) dx<\/p>\n
\\(\\int\\) \\(1\\over v^2 + 2v + 1- 1\\) dv = \\(\\int\\) \\(-1\\over x\\) dx<\/p>\n
\\(\\int\\) \\(1\\over (v + 1)^2 – 1^2\\) dv = \\(\\int\\) \\(-1\\over x\\) dx<\/p>\n
\\(1\\over 2\\) \\(log{{v+1 – 1}\\over {v+1+1}}\\) = -log x + log C<\/p>\n
\\(1\\over 2\\) \\(log{{v}\\over {v+2}}\\) = -log x + log C<\/p>\n
\\(log{{v}\\over {v+2}}\\) + 2log x = 2log C<\/p>\n
\\(log|{vx^2\\over v+2}|\\) = log k<\/p>\n
Put v = y\/x<\/p>\n
k = \\(x^2y\\over y + 2x\\), which is the solution of differential equation.<\/p>\n\n\n