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
372 views
in Olympiad by (30 points)
edited by

Let \( y=y(x), x>1 \), be the solution of the differential equation \( (x-1) \frac{d y}{d x}+2 x y=\frac{1}{x-1} \), with \( y(2)=\frac{1+e^{4}}{2 e^{4}} \). If \( y(3)=\frac{e^{\alpha}+1}{\beta e^{\alpha}} \). then the value of \( \alpha+\beta \) is equal to

Please log in or register to answer this question.

1 Answer

0 votes
by (53.3k points)

\((x - 1) \frac{dy}{dx} + 2xy = \frac 1{x - 1}\)

⇒ \(\frac{dy}{dx} + \frac{2x}{x -1}y = \frac1{(x - 1)^2}\)

which is a linear differential equation

\(\therefore I.F. = e^{\int pdx} = e^{\int \frac{2x}{x-1}dx}\)

\(= e^{\int \frac{2x -2 +2}{x -1}dx}\)

\(= e^{\int (2 + \frac 2{x-1})dx}\)

\(= e^{2x + 2log\,(x -1)}\)

\(= e^{2x}.e^{log(x -1)^2}\)      \((\because e^{x + y} = e^x.e^y)\)

\(= (x - 1)^2 e^{2x}\)        \((\because e^{log(f(x))} = f(x))\)

\(\therefore \) Common solution is 

\(y \times I.F. = \int Q \times (I.F.) dx + C\)

⇒ \(y(x - 1)^2 e^{2x} = \int \frac 1{(x - 1)^2} \times e^{2x} (x - 1)^2 dx +C\)

⇒ \(y (x -1 )^2 e^{2x} = \frac{e^{2x}}2 + C\)     ......(i)

\(\because y(2) = \frac{1+e^4}{2e^4}\)    (given)

\(\therefore \) By putting x = 2 in equation (i), we obtain 

\(\frac{1 + e^4}{2e^4} \times 1 \times e^4 = \frac{e^4}2 + C\)

⇒ \(C = \frac{1 + e^4}2 - \frac{e^4}2 = \frac 12\)

\(\therefore \) From (i), we get

\(y(x - 1)^2 e^{2x} = \frac{e^{2x}}2 + \frac12\)

⇒ \(y = \frac{e^{2x} + 1}{2(x - 1)^2 e^{2x}}\)    ......(ii)

⇒ \(y(3) = \frac{e^{6x}+ 1}{2(3 - 1)^2 e^{6x}}\)    (By putting x = 3 in equation (ii))

⇒ \(y(3) = \frac{e^{6x} + 1}{8 \,e^{6x}}\)

\(\therefore \alpha = 6, \beta = 8\)     \((\because y = \frac{e^\alpha+1}{\beta \,e^\alpha}(given))\)

\(\therefore \alpha + \beta = 6 + 8 = 14\)

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

...