JEE MAIN - Mathematics (2016 - 10th April Morning Slot - No. 9)
For x $$ \in $$ R, x $$ \ne $$ 0, if y(x) is a 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 :
(where C is a constant.)
x $$\int\limits_1^x y $$ (t) dt = (x + 1) $$\int\limits_1^x ty $$ (t) dt, then y (x) equals :
(where C is a constant.)
$${C \over x}{e^{ - {1 \over x}}}$$
$${C \over {{x^2}}}{e^{ - {1 \over x}}}$$
$${C \over {{x^3}}}{e^{ - {1 \over x}}}$$
$$C{x^3}\,{1 \over {{e^x}}}$$
Explanation
$$x\int\limits_1^x {y\left( t \right)dt} = x\int\limits_1^x {ty} \left( t \right)dt + \int\limits_1^x {ty\left( t \right)} \,\,dt$$
Differentiate w.r. to x.
$$\int\limits_1^x {y\left( t \right)dt + x\left[ {y\left( x \right) - y\left( 1 \right)} \right]} $$
$$ = \int\limits_1^x {ty\left( t \right)dt + x\left[ {xy\left( x \right) - y\left( 1 \right)} \right] + xy\left( x \right) - y\left( 1 \right)} $$
$$\int\limits_1^x {y\left( t \right)dt = \int\limits_1^x {ty\left( t \right)dt + {x^2}y\left( x \right) - y\left( 1 \right)} } $$
Diff. again w.r.t to x
$$y\left( x \right) - y\left( a \right) = xy\left( x \right) - y\left( a \right) + 2x\,y\left( x \right) + {x^2}{y^3}\left( x \right)$$
$$\left( {1 - 3x} \right)y\left( x \right) = {x^2}{y^3}\,\left( x \right)$$
$${{{y^3}\left( x \right)} \over {y\left( x \right)}} = {{1 - 3x} \over {{x^2}}}$$
$${{1dy} \over {ydx}} = {{1 - 3x} \over {{x^2}}} \Rightarrow \ln \,y = - {1 \over x} - 3\ln x$$
$$\ln \left( {y{x^3}} \right) = - {1 \over x}$$
$$y{x^3} = - {e^{ - 1/x}}$$
$$y = {{{e^{ - {1 \over x}}}} \over {{x^3}}}$$
or $$y = {{c{e^{ - {1 \over x}}}} \over {{x^3}}}$$
Differentiate w.r. to x.
$$\int\limits_1^x {y\left( t \right)dt + x\left[ {y\left( x \right) - y\left( 1 \right)} \right]} $$
$$ = \int\limits_1^x {ty\left( t \right)dt + x\left[ {xy\left( x \right) - y\left( 1 \right)} \right] + xy\left( x \right) - y\left( 1 \right)} $$
$$\int\limits_1^x {y\left( t \right)dt = \int\limits_1^x {ty\left( t \right)dt + {x^2}y\left( x \right) - y\left( 1 \right)} } $$
Diff. again w.r.t to x
$$y\left( x \right) - y\left( a \right) = xy\left( x \right) - y\left( a \right) + 2x\,y\left( x \right) + {x^2}{y^3}\left( x \right)$$
$$\left( {1 - 3x} \right)y\left( x \right) = {x^2}{y^3}\,\left( x \right)$$
$${{{y^3}\left( x \right)} \over {y\left( x \right)}} = {{1 - 3x} \over {{x^2}}}$$
$${{1dy} \over {ydx}} = {{1 - 3x} \over {{x^2}}} \Rightarrow \ln \,y = - {1 \over x} - 3\ln x$$
$$\ln \left( {y{x^3}} \right) = - {1 \over x}$$
$$y{x^3} = - {e^{ - 1/x}}$$
$$y = {{{e^{ - {1 \over x}}}} \over {{x^3}}}$$
or $$y = {{c{e^{ - {1 \over x}}}} \over {{x^3}}}$$
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