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

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.