Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
129 views
in Differential Equations by (20 points)
Solve \( x \frac{d y}{d x}+\frac{2 y}{x}=\frac{1}{x^{3}} \)

Please log in or register to answer this question.

1 Answer

0 votes
by (45.0k points)

x\(\frac{dy}{dx}+\frac{2y}x=\frac1{x^3}\)

⇒ \(\frac{dy}{dx}+\frac{2}{x^2}y=\frac1{x^4}\) which is linear differential equation.

∴ I.F. = \(e^{\int\frac2{x^2}dx} = e^{\frac{-2}x}\)

y x I.F. = \(\int\frac1{x^4}(I.F)dx\)

⇒ y. e\(\frac{-2}x\) = \(\frac{e^{\frac2x}}{x^4}dx\)

Let \(\frac{-2}x=t\)

⇒ \(\frac2{x^2}dx=dt\)

∴ ye\(\frac{-2}x\)  = \(\frac12\int\frac{t^2}4e^tdt\)

 = \(\frac18\)(t2et - 2tet + et) + c

 = \(\frac18\)(t2 - 2t + 1)et + c

left part

y.e-2/x = \(\frac18\)(t - 1)2et + c

⇒ ye-2/x = \(\frac18\) (-2/x - 1)2e-2/x + c ( ∵ t = -2/x)

⇒ y = \(\frac18\)(2/x + 1)2 + ce2/x which is general solution of given differential equation.

No related questions found

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...