The given differential equation is:<\/p>\n\n\n\n
[
x^3 y” + 5x^2 y’ + 7xy’ – 7y = 0
]<\/p>\n\n\n\n
Step 1: Recognize the Type of Equation<\/h3>\n\n\n\n
This is a linear second-order homogeneous differential equation with variable coefficients. We can try to solve it using the method of Cauchy-Euler equation<\/strong> since it has powers of (x) multiplying the derivatives. This suggests the possibility of a solution of the form:<\/p>\n\n\n\n [ where (r) is a constant to be determined.<\/p>\n\n\n\n First, compute the necessary derivatives:<\/p>\n\n\n\n Now, substitute these into the original differential equation:<\/p>\n\n\n\n [ Simplify each term:<\/p>\n\n\n\n [ Now, combine like terms:<\/p>\n\n\n\n [ This leads to the equation:<\/p>\n\n\n\n [ Simplify the terms:<\/p>\n\n\n\n [ For this equation to hold for all values of (x), both terms must independently sum to zero. This gives us two separate equations:<\/p>\n\n\n\n [ [ Since we have three possible values for (r): (r = 0), (r = -4), and (r = 1), the general solution is a linear combination of the corresponding solutions:<\/p>\n\n\n\n [ Thus, the general solution is:<\/p>\n\n\n\n [ In this approach, we used the method of solving a Cauchy-Euler equation by assuming a solution of the form (y = x^r), and then solved for the possible values of (r). By substituting these values into the general solution form, we obtained the three independent solutions, which we combined to form the general solution. The constant coefficients (C_1), (C_2), and (C_3) are arbitrary and depend on the initial conditions or boundary conditions if provided.<\/p>\n\n\n\n I will now provide an image representation of the solution process.<\/p>\n\n\n\n Here is the diagram illustrating the solution process for the differential equation. It shows the substitution, differentiation steps, and how we derive the general solution ( y(x) = C_1 + C_2 x^{-4} + C_3 x ). Let me know if you need further clarifications or explanations!<\/p>\n","protected":false},"excerpt":{"rendered":" Find the general solution of the differential equation: x3y”+ 5x2y”&39; + 7xy&39;-7y=0. The Correct Answer and Explanation is : The given differential equation is: [x^3 y” + 5x^2 y’ + 7xy’ – 7y = 0] Step 1: Recognize the Type of Equation This is a linear second-order homogeneous differential equation with variable coefficients. We can […]<\/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-196536","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/196536","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=196536"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/196536\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=196536"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=196536"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=196536"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
y = x^r
]<\/p>\n\n\n\nStep 2: Substitute (y = x^r) into the Differential Equation<\/h3>\n\n\n\n
\n
x^3 \\cdot r(r-1) x^{r-2} + 5x^2 \\cdot r x^{r-1} + 7x \\cdot r x^{r-1} – 7 x^r = 0
]<\/p>\n\n\n\n
r(r-1) x^{r+1} + 5r x^{r+1} + 7r x^{r} – 7 x^r = 0
]<\/p>\n\n\n\n
(r(r-1) + 5r) x^{r+1} + (7r – 7) x^r = 0
]<\/p>\n\n\n\n
[(r^2 – r + 5r)] x^{r+1} + (7r – 7) x^r = 0
]<\/p>\n\n\n\n
(r^2 + 4r) x^{r+1} + (7r – 7) x^r = 0
]<\/p>\n\n\n\nStep 3: Set up the Indicial Equation<\/h3>\n\n\n\n
\n
Solving the First Equation:<\/h4>\n\n\n\n
r^2 + 4r = 0 \\implies r(r + 4) = 0
]
Thus, (r = 0) or (r = -4).<\/p>\n\n\n\nSolving the Second Equation:<\/h4>\n\n\n\n
7r – 7 = 0 \\implies r = 1
]<\/p>\n\n\n\nStep 4: Construct the General Solution<\/h3>\n\n\n\n
y(x) = C_1 x^0 + C_2 x^{-4} + C_3 x^1
]<\/p>\n\n\n\n
y(x) = C_1 + C_2 x^{-4} + C_3 x
]<\/p>\n\n\n\nExplanation:<\/h3>\n\n\n\n
![\"\"](\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/03\/image-218.png\")
<\/figure>\n\n\n\n