QJEEM16Ma8abe: JEE Main 2016 Mathematics - Class 12 JEEM16Ma8abe Solution: For x R , x 0, if y(x) is a differentiable function such that x …

Question

Verified Solution • 2025–26 Session moderate

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.)

Accepted Answer

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}}}

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