For x R, x 0, if y(x) is a differentiable function such that x_1^x y(t),dt = (x+1)_1^x ty(…

Question

For $x \in R$, $x 
eq 0$, if $y(x)$ is a differentiable function such that $x\displaystyle\int_1^x y(t)\,dt = (x+1)\int_1^x ty(t)\,dt$, 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   $$y = {{c{e^{ - {1 \over x}}}} \over {{x^3}}}$$

Solutions · 1