Question

For \(x\in\mathbf{R}\), \(x\neq0\), if \(y(x)\) is a differentiable function such that \(x\displaystyle\int_1^xy(t)\,dt=(x+1)\displaystyle\int_1^xt\,y(t)\,dt\), then \(y(x)\) equals :
(where C is a constant.)

• A. \(\frac{C}{x}e^{-\frac{1}{x}}\)
• B. \(\frac{C}{x^2}e^{-\frac{1}{x}}\)
• C. \(\frac{C}{x^3}e^{-\frac{1}{x}}\)
• D. \(Cx^3e^{\frac{1}{x}}\)

Hint:

Differentiate the relation twice to remove the integrals and obtain \(x^2y'+(3x-1)y=0\), then separate variables.

Answer 1

The Correct Option is C

Differentiating \(x\int_1^xy\,dt=(x+1)\int_1^xty\,dt\): \(\int_1^xy\,dt+xy=\int_1^xty\,dt+(x+1)xy\).
Differentiate again: \(y+y+xy'=xy+(2x+1)y+(x^2+x)y'\).
\(\Rightarrow x^2y'+(3x-1)y=0\Rightarrow\frac{y'}{y}=-\frac{3}{x}+\frac{1}{x^2}\).
Integrate: \(\ln y=-3\ln x-\frac{1}{x}+c\), so \(y=\frac{C}{x^3}e^{-\frac{1}{x}}\).
Answer: (3) \(\frac{C}{x^3}e^{-\frac{1}{x}}\)