Question

For $x \in R$,$x \ne 0$, $y$ is differentiable function such that $x\int\limits_1^x {y(t)dt} $$ = (x + 1)$$\int\limits_1^x {ty(t)dt} $, then $y(x)$equals:
A. $C{x^3}{e^{\dfrac{1}{x}}}$
B. $\dfrac{C}{{{x^2}}}{e^{\dfrac{{ - 1}}{x}}}$
C. $\dfrac{C}{x}{e^{\dfrac{{ - 1}}{x}}}$
D. $\dfrac{C}{{{x^3}}}{e^{\dfrac{{ - 1}}{x}}}$

Answer 1

Hint: In this question, we are given that $x \in R$,$x \ne 0$, $y$ is a differentiable function. Here we should take differential with respect to $x$ of
$x\int\limits_1^x {y(t)dt} $$ = (x + 1)$$\int\limits_1^x {ty(t)dt} $ and then we will find $y(x)$ by this equation.

Complete step-by-step answer:
We are given that
$x \in R$,$x \ne 0$, $y$ is differentiable function such that $x\int\limits_1^x {y(t)dt} $$ = (x + 1)$$\int\limits_1^x {ty(t)dt} $
We can also write it as
\[\]$x\int\limits_1^x {y(t)dt} $$ = x$$\int\limits_1^x {ty(t)dt} $$ + \int\limits_1^x {ty(t)dt} $$ - - - - - - (1)$
As we have to find $y(x)$ we have to proceed with (1),
Taking differential of both the sides with respect to $x$ of (1)
$\dfrac{d}{{dx}}(x\int\limits_1^x {y(t)dt)} $$ = \dfrac{d}{{dx}}x$$\int\limits_1^x {ty(t)dt} $$ + \dfrac{d}{{dx}}\int\limits_1^x {ty(t)dt} $
Now using the product rule,
$\dfrac{d}{{dx}}(u.v) = v\dfrac{{du}}{{dx}} + u\dfrac{{dv}}{{dx}}$
We get,
$\int\limits_1^x {y(t)dt.\dfrac{{dx}}{{dx}} + x\dfrac{{d\int\limits_1^x {y(t)dt} }}{{dx}} = \int\limits_1^x {ty(t)dt\dfrac{{dx}}{{dx}}} } + \dfrac{d}{{dx}}$$\int\limits_1^x {ty(t)dt} $$.x$$ + \dfrac{d}{{dx}}\int\limits_1^x {ty(t)dt} $
Now we know that $\dfrac{{dx}}{{dx}} = 1,\dfrac{{d\int {f(y)} }}{{dx}} = f(y)$
$\int\limits_1^x {y(t)dt + x(y(x) - y(1)) = \int\limits_1^x {ty(t)dt} } $$ + x(xy(x) - 1.y(1))$$ + (xy(x) - 1.y(1))$
$\int\limits_1^x {y(t)dt + xy(x) - xy(1) = \int\limits_1^x {ty(t)dt} } $$ + {x^2}y(x) - xy(1))$$ + xy(x) - 1.y(1)$
$\int\limits_1^x {y(t)dt} $$ = \int\limits_1^x {ty(t)dt} $$ + {x^2}y(x) - y(1)$$ - - - - - (2)$
Taking differential of both the sides with respect to $x$of (2)
$\dfrac{d}{{dx}}\int\limits_1^x {y(t)dt} $$ = \dfrac{d}{{dx}}\int\limits_1^x {ty(t)dt} $$ + \dfrac{d}{{dx}}{x^2}y(x) - \dfrac{d}{{dx}}y(1)$
Now using the product rule again,
 $\dfrac{d}{{dx}}(u.v) = v\dfrac{{du}}{{dx}} + u\dfrac{{dv}}{{dx}}$
$\dfrac{{dc}}{{dx}} = 0$, where $c$ is any constant
We get,
$y(x) - y(1) = xy(x) - y(1) + 2xy(x)$$ + {x^2}\dfrac{d}{{dx}}y(x)$
$y(x) - 3xy(x)$$ = {x^2}\dfrac{d}{{dx}}y(x)$
$\dfrac{1}{{y(x)}}.\dfrac{d}{{dx}}y(x)$$ = \dfrac{{1 - 3x}}{{{x^2}}}$
$\dfrac{1}{{y(x)}}.dy(x)$$ = \dfrac{{1 - 3x}}{{{x^2}}}.dx$$ - - - - - (3)$
Now integrating both sides of (3), we use
$\int {\dfrac{1}{{y(x)}}.dy(x)} $$ = \int {\dfrac{1}{{{x^2}}} - \dfrac{3}{x}.dx} $
Now we know that $\ln x + {c_1} = \int {\dfrac{1}{x}.dx} $and
$\int {{x^n}.dx} = \dfrac{{{x^{n + 1}}}}{{n + 1}} + {c_2}$
So, we get:
$\ln y(x) = \dfrac{{{x^{ - 1}}}}{{ - 1}} - 3\ln x + {c_2} - {c_1}$
Now putting ${c_3} = {c_2} - {c_1}$
Where ${c_3}$is the constant, we get
$\ln y(x) + 3\ln x = \dfrac{{{x^{ - 1}}}}{{ - 1}} + \ln {c_3}$
We know that
$a\ln b = \ln {b^a}$
So, we get
$\ln y(x) + \ln {x^3} = \dfrac{{{x^{ - 1}}}}{{ - 1}} + \ln {c_3}$
Now taking exponential on both sides we get,
${e^{\ln y(x){x^3}}} = {e^{ - \dfrac{1}{x} + \ln {c_3}}}$
Now using ${e^{\ln a}} = a$ so we get,
$y(x){x^3} = {e^{\dfrac{{ - 1}}{x} + \ln {c_3}}}$
Now using ${e^{a + b}} = {e^a}{e^b}$
$y(x){x^3} = {e^{ - \dfrac{1}{x}}} + {e^{\ln {c_3}}}$
Now again using ${e^{\ln a}} = a$, we get,
$y(x){x^3} = {e^{\dfrac{{ - 1}}{x}}}.{c_3}$
$y(x) = \dfrac{1}{{{x^3}}}{e^{\dfrac{{ - 1}}{x}}}.{c_3}$
Now we can put $c = {c_3}$ as it is just a constant
$y(x) = \dfrac{c}{{{x^3}}}{e^{\dfrac{{ - 1}}{x}}}$

So, the correct answer is “Option D”.

Note: In this question, we are writing
$\dfrac{d}{{dx}}\int\limits_1^x {y(t)dt} $$ = y(x) - y(1).................(A)$
because we know that:
$\int\limits_a^b {f(x)dx} = \left[ {g(x)} \right]_a^b = g(b) - g(a)$
Where
$\int {f(x)dx} = \left[ {g(x)} \right]$
And proof for (A) is,
Let $\int\limits_1^x {y(t)dt} = z(t)$
Now, we know that if $\int {f(x)dx} = g(x)$, then
$\dfrac{{dg(x)}}{{dx}} = f(x)$
So, using this formula, we can say that
$
  \dfrac{d}{{dx}}\int\limits_1^x {y(t)dt} = \dfrac{{dz(x)}}{{dx}} - \dfrac{{dz(1)}}{{dx}} \\
  {\text{ }} = y(x) - y(1) \\
 $
Hence this is proof for (A) which is used in the above question and is an important part to solve the question.