Question
The solution of the equation $$x\int_0^x {y\left( t \right)dt = \left( {x + 1} \right)\int_0^x {ty\left( t \right)dt,\,x > 0} } {\text{ is :}}$$
A.
$$y = \frac{c}{{{x^3}}}{e^{{x^3}}}$$
B.
$$y = c{x^3}{e^{ - {x^3}}}$$
C.
$$\frac{c}{{{x^3}}}{e^{ - x}}$$
D.
none of these
Answer :
none of these
Solution :
The equation is
$$x\int_0^x {y\left( t \right)dt = \left( {x + 1} \right)} \int_0^x {ty\left( t \right)dt......\left( {\text{i}} \right)} $$
Differentiating both the sides with respect to $$x,$$ we get
$$\eqalign{
& xy\left( x \right) + \int_0^x {y\left( t \right)dt = \left( {x + 1} \right)xy\left( x \right)} + \int_0^x {ty\left( t \right)dt} \cr
& \Rightarrow \int_0^x {y\left( t \right)dt = {x^2}y\left( x \right)} + \int_0^x {ty\left( t \right)dt......\left( {{\text{ii}}} \right)} \cr} $$
Differentiating again with respect to $$x,$$ we get
$$\eqalign{
& y\left( x \right) = 2xy\left( x \right) + {x^2}y'\left( x \right) + xy\left( x \right) \cr
& \Rightarrow {x^2}\frac{{dy}}{{dx}} = \left( {1 - 3x} \right)y\,\,\,\,\,\,\,\,\left[ {{\text{writing }}y\left( x \right) = y} \right] \cr
& \Rightarrow \frac{{dy}}{y} = \left( {\frac{{1 - 3x}}{{{x^2}}}} \right)dx \cr
& \Rightarrow \frac{{dy}}{y} = \left( {\frac{1}{{{x^2}}} - \frac{3}{x}} \right)dx \cr} $$
Integrating we get, $$\log \,y = - \frac{1}{x} - 3\,\log \,x + a,\,\,\,a$$ is constant
$$\eqalign{
& \Rightarrow \log \,y + 3\,\log \,x = a - \frac{1}{x} \cr
& \Rightarrow \log \left( {y{x^3}} \right) = a - \frac{1}{x} \cr
& \Rightarrow y{x^3} = {e^{a - \frac{1}{x}}} \cr
& \Rightarrow y{x^3} = c.{e^{ - \frac{1}{x}}}{\text{ where }}c = {e^a} \cr
& \therefore \,y = \frac{c}{{{x^3}}}{e^{ - \frac{1}{x}}} \cr} $$