Question

Let y = y(x) be the solution of differential equation, $\sf x\dfrad{dy}{dx} + y = xlog_ex$ (x > 1). If $\sf 2y(2) = log_e4 -1$. Then y(e) =
1) e²/4
2) e/4
3) -e/2
d) -e²/2​

Answer 1

GIVEN :–

• Differential equation –

$\\ \implies \bf x\dfrac{dy}{dx} + y = xlog_ex$

• And –

$\\ \bf \implies 2y(2) = log_e(4) -1$

TO FIND :–

• Value of y(e) = ?

SOLUTION :–

$\\ \implies \bf x\dfrac{dy}{dx} + y = xlog_ex$

$\\ \implies \bf \dfrac{dy}{dx} + \dfrac{y}{x}= log_ex$

• It's a linear differential equation. Now let's compare with –

$\\ \implies \bf \dfrac{dy}{dx} +Py= Q$

• So –

$\\ \implies \bf P = \dfrac{1}{x} \: \: \: and \: \: \: Q = log_ex$

• Now Let's find I.F. –

$\\ \implies \bf I.F. = e^{\int P.dx}$

$\\ \implies \bf I.F. = e^{\int \frac{1}{x} .dx}$

$\\ \implies \bf I.F. = e^{ log_{e}(x) }$

$\\ \implies \large{ \boxed{\bf I.F. = x}}$

• SOLUTION OF DIFFERENTIAL EQUATION :–

$\\ \large \implies{ \boxed{ \bf y(I.F.) = \int(I.F.)Q.dx \: + c}}$

$\\\implies \bf y(x) = \int(x)log_ex.dx \: + c$

• Using identity –

$\\ \large\implies{ \boxed{ \bf \int(u.v).dx = u \int v.dx - \int \bigg(\dfrac{du}{dx}. \int v.dx \bigg).dx}}$

• So that —

$\\\implies \bf y(x) = log_ex\int x.dx \: - \int \bigg(\dfrac{d(log_ex)}{dx}. \int x.dx \bigg).dx + c$

$\\\implies \bf y(x) =(log_ex) \dfrac{ {x}^{2} }{2} - \int \dfrac{1}{x} . \dfrac{ {x}^{2} }{2} .dx + c$

$\\\implies \bf y(x) =(log_ex) \dfrac{ {x}^{2} }{2} - \int \dfrac{ {x}}{2} .dx + c$

$\\\implies \large{ \boxed{ \bf y(x) =(log_ex) \dfrac{ {x}^{2} }{2} - \dfrac{ {x}^{2} }{4} + c}}$

• And –

$\\ \bf \implies 2y(2) = log_e(4) -1$

$\\ \bf \implies y(2) = \dfrac{log_e(4) -1}{2}$

• Now put x = 2 –

$\\\implies \large \bf (2) \bigg(\dfrac{log_e(4) -1}{2}\bigg) =(log_e(2)) \dfrac{ {2}^{2} }{2} - \dfrac{ {2}^{2} }{4} + c$

$\\\implies \large \bf log_e(4) -1 =(log_e(2)) \dfrac{4}{2} - \dfrac{4}{4} + c$

$\\\implies \large \bf log_e(2^2) -1 =(log_e(2))2-1+ c$

$\\\implies \large \bf 2log_e(2) -1 =2log_e(2)-1+ c$

$\\\longrightarrow \large {\boxed { \bf c=0}}$

• So that –

$\\\implies \large \bf y(x) =(log_ex) \dfrac{ {x}^{2} }{2} - \dfrac{ {x}^{2} }{4} + 0$

• Now put x = e –

$\\\implies \large \bf y(e)=(log_e{e}) \dfrac{ {e}^{2} }{2} - \dfrac{ {e}^{2} }{4}$

$\\\implies \large \bf y(e)= \dfrac{ {e}^{2} }{2} - \dfrac{ {e}^{2} }{4}$

$\\\implies \large \bf y(e)=\dfrac{ {e}^{2} }{4}$

$\\\implies \large {\boxed{\bf y=\dfrac{e}{4}}}$

▪︎ Hence, Option (b) is correct.

Answer 2

$\huge\bf{\underline{\underline{\gray{GIVEN:-}}}}$

✞︎ y = y(x) be the solution of the differential equation,

• $\bf \blue {x\dfrac{dy}{dx} + y = xlog_ex}$

• $\bf\gray{And \: 2y(2) = log_e4 - 1}$

$\huge\bf{\underline{\underline{\gray{TO\:FIND:-}}}}$

• The value of y(e) .

$\huge\bf{\underline{\underline{\gray{SOLUTION:-}}}}$

Given that,

• $\bf {x\dfrac{dy}{dx} + y = xlog_ex}$

☞︎︎︎ Divide the above equation by x .

$\rm {\implies \dfrac{x\dfrac{dy}{dx}}{x} + \dfrac{y}{x} = \dfrac{xlog_ex}{x}}$

$\rm {\implies \dfrac{dy}{dx} + \dfrac{1}{x}y = log_ex}$

☞︎︎︎ We know that, the standard form of the Linear differential equation is

$\huge\red\checkmark$ $\bf \orange {\dfrac{dx}{dy} + p(x) . y = q(x)}$

Where,

• $\bf \red {p(x)}$ = $\bold {\dfrac{1}{x}}$

• $\bf \red {q(x)}$ = $\bold {log_ex = lnx}$

☞︎︎︎ Here Integrating factor (I.F) is,

$\purple \bigstar$ $\bf \red {I.F = e^{\int {p(x) . dx}}}$

$\rm {\implies I.F = e^{\int \frac{1}{x} . dx}}$

$\rm {\implies I.F = e^{ln x}}$

$\bf {\implies I.F = x}$

☞︎︎︎ After solving the above differential equation, we get

• $\bf \pink {y . (I.F) = \int \Big({q(x) . (I.F)}\Big) dx}$

$\rm {\implies y . x = \int {\Big(lnx . (x)\Big)} dx}$

$\rm {\implies y . x = \int {\Big(x . lnx\Big)} dx}$

☞︎︎︎ Now using By parts method,

$\bf \green {\int {(u . v)} dx = u \int {v} dx - \displaystyle {\int \Big(\dfrac{du}{dx} . \int {v} . dx \Big)} dx}$

$\rm {\implies xy = lnx \int {x} dx - \displaystyle {\int \Big(\dfrac{d(lnx)}{dx} . \int {x} . dx \Big)} dx}$

$\rm {\implies xy = \dfrac{x^2}{2} lnx - \displaystyle \dfrac{1}{2} \int \Big({\dfrac{1}{x} . x^2 \Big)} dx}$

$\rm {\implies xy = \dfrac{x^2}{2} lnx - \displaystyle \dfrac{1}{2} \int {x} . dx}$

$\rm {\implies xy = \dfrac{x^2}{2} lnx - \displaystyle \dfrac{1}{2} . \dfrac{x^2}{2} + c}$

$\bf {\implies xy = \dfrac{x^2}{2} lnx - \dfrac{x^2}{4} + c}$

__________________________

☯︎ According to the question,

$\red \bullet$ $\bf \gray {2y(2) = log_e4 - 1}$

✯ If x = 2, $\rm {\implies y(2) = \dfrac{ln4 - 1}{2}}$

$\rm {\implies 2 . \dfrac{ln4 - 1}{2} = \dfrac{2^2}{2} ln2 - \displaystyle \dfrac{2^2}{4} + c}$

$\rm {\implies ln4 - 1 = 2 ln2 - 1 + c}$

$\rm {\implies ln4 - 1 = ln{(2^2)} - 1 + c}$

$\rm {\implies \cancel{ln4 - 1} = \cancel{ln4 - 1} + c}$

$\bf \blue {\implies c = 0}$

__________________________

☞︎︎︎ Putting the value of 'c = 0' in the above equation, we get

$\bf \gray {\implies xy = \dfrac{x^2}{2} lnx - \dfrac{x^2}{4} + 0}$

$\rm {\implies xy = x \Big(\dfrac{x}{2} lnx - \dfrac{x}{4}\Big)}$

$\rm {\implies y = \Big(\dfrac{x}{2} lnx - \dfrac{x}{4}\Big)}$

Now,

$\rm {\implies y(e) = \Big(\dfrac{e}{2} lne - \dfrac{e}{4}\Big)}$

• ln(e) = 1

$\rm {\implies y(e) = \Big(\dfrac{e}{2} - \dfrac{e}{4}\Big)}$

$\rm {\implies y(e) = \Big(\dfrac{2e - e}{4}\Big)}$

$\bf \purple {\implies y(e) = \dfrac{e}{4}}$

$\huge\red\therefore$ (b) The value of y(e) is $\bf {\dfrac{e}{4}}$ .