{"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

Solution of Homogeneous Differential Equation<\/h2>\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

\n

1). 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

\n
Next – Solution of Linear Differential Equation<\/a><\/div>\n<\/div>\n\n\n\n
\n
Previous – Differential Equations Reducible to Variable Separable Form<\/a><\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

Here you will learn how to find solution of homogeneous differential equation of first order first degree with examples. Let’s begin – Solution of Homogeneous Differential Equation If a first degree first order differential equation is expressible in the form \\(dy\\over dx\\) = \\(f(x, y)\\over g(x, y)\\) where f(x, y) and g(x, y) are homogeneous …<\/p>\n

Solution of Homogeneous Differential Equation<\/span> Read More »<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"default","ast-global-header-display":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"default","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":""},"categories":[38],"tags":[251,254,252,253],"yoast_head":"\nSolution of Homogeneous Differential Equation - Mathemerize<\/title>\n<meta name=\"description\" content=\"In this post you will learn how to find solution of homogeneous differential equation of first order first degree with examples.\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/mathemerize.com\/solution-of-homogeneous-differential-equation\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Solution of Homogeneous Differential Equation - 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