Solve The Separable Differential Equation Dx\/Dt = X^2 + 1\/49, And Find The Particular Solution Satisfying The Initial Condition X(0) = -2.<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n It seems that the image you provided isn’t accessible in this environment, but based on your description of the separable differential equation:<\/p>\n\n\n\n [ Let’s proceed to solve it step by step and find the particular solution for the initial condition ( x(0) = -2 ).<\/p>\n\n\n\n We begin by separating the variables (x) and (t). To do this, we rewrite the equation as:<\/p>\n\n\n\n [ The left-hand side involves a rational function in terms of (x). First, we factor out ( \\frac{1}{49} ) from the denominator:<\/p>\n\n\n\n [ So the equation becomes:<\/p>\n\n\n\n [ Simplifying further:<\/p>\n\n\n\n [ Now, we integrate both sides. The integral on the left-hand side is a standard form that can be handled using a standard trigonometric substitution. The integral is:<\/p>\n\n\n\n [ This is a standard integral form where:<\/p>\n\n\n\n [ In our case, ( a = \\frac{1}{7} ), so:<\/p>\n\n\n\n [ Thus, we now have:<\/p>\n\n\n\n [ Now, we use the initial condition ( x(0) = -2 ) to find the value of the constant ( C ). Substituting ( t = 0 ) and ( x = -2 ) into the equation:<\/p>\n\n\n\n [ Simplifying:<\/p>\n\n\n\n [ So, ( C = 7 \\arctan(-14) ).<\/p>\n\n\n\n Thus, the general solution is:<\/p>\n\n\n\n [ To express ( x ) explicitly in terms of ( t ), we solve for ( x ):<\/p>\n\n\n\n [ [ [ This is the particular solution satisfying the initial condition ( x(0) = -2 ).<\/p>\n\n\n\n The solution to the given differential equation, satisfying the initial condition, is:<\/p>\n\n\n\n [ This solution describes how ( x ) evolves over time based on the given equation.<\/p>\n","protected":false},"excerpt":{"rendered":" Solve The Separable Differential Equation Dx\/Dt = X^2 + 1\/49, And Find The Particular Solution Satisfying The Initial Condition X(0) = -2. The Correct Answer and Explanation is : It seems that the image you provided isn’t accessible in this environment, but based on your description of the separable differential equation: [\\frac{dx}{dt} = x^2 + […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","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":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-188773","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/188773","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=188773"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/188773\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=188773"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=188773"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=188773"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\\frac{dx}{dt} = x^2 + \\frac{1}{49}
]<\/p>\n\n\n\nStep 1: Separate the variables<\/h3>\n\n\n\n
\\frac{dx}{x^2 + \\frac{1}{49}} = dt
]<\/p>\n\n\n\nStep 2: Simplify the left-hand side<\/h3>\n\n\n\n
x^2 + \\frac{1}{49} = \\frac{49x^2 + 1}{49}
]<\/p>\n\n\n\n
\\frac{dx}{\\frac{49x^2 + 1}{49}} = dt
]<\/p>\n\n\n\n
\\frac{49dx}{49x^2 + 1} = dt
]<\/p>\n\n\n\nStep 3: Integrate both sides<\/h3>\n\n\n\n
\\int \\frac{49}{49x^2 + 1} dx = \\int dt
]<\/p>\n\n\n\n
\\int \\frac{dx}{a^2 + x^2} = \\frac{1}{a} \\arctan\\left(\\frac{x}{a}\\right)
]<\/p>\n\n\n\n
\\int \\frac{49}{49x^2 + 1} dx = 7 \\arctan(7x)
]<\/p>\n\n\n\n
7 \\arctan(7x) = t + C
]<\/p>\n\n\n\nStep 4: Solve for the constant of integration<\/h3>\n\n\n\n
7 \\arctan(7(-2)) = 0 + C
]<\/p>\n\n\n\n
7 \\arctan(-14) = C
]<\/p>\n\n\n\nFinal solution<\/h3>\n\n\n\n
7 \\arctan(7x) = t + 7 \\arctan(-14)
]<\/p>\n\n\n\n
\\arctan(7x) = \\frac{t}{7} + \\arctan(-14)
]<\/p>\n\n\n\n
7x = \\tan\\left(\\frac{t}{7} + \\arctan(-14)\\right)
]<\/p>\n\n\n\n
x = \\frac{1}{7} \\tan\\left(\\frac{t}{7} + \\arctan(-14)\\right)
]<\/p>\n\n\n\nConclusion:<\/h3>\n\n\n\n
x(t) = \\frac{1}{7} \\tan\\left(\\frac{t}{7} + \\arctan(-14)\\right)
]<\/p>\n\n\n\n