If f(x) satisfies the relation f(x)=e^x+ _1^0(y+xe^x)f(y)dy, then e+f(0) is equal to ____ …

Question

If $f(x)$ satisfies the relation $f(x)={e}^{x}+{\int }_{1}^{0}(y+x{e}^{x})f(y)dy$, then $e+f(0)$ is equal to $____$ .

Answer: 2

Accepted Answer

$\begin{gathered} f(x)={e}^{x}+{\int }_{1}^{0}(y+x{e}^{x})f(y)dy \ f(x)={e}^{x}+{\int }_{1}^{0}yf(y)dy+{\int }_{1}^{0}x{e}^{x}f(y)dy \end{gathered}$

Integration is w.r.t $y$, so variable $x$ is taken as a constant.

$f(x)={e}^{x}+{\int }_{1}^{0}yf(y)dy+x{e}^{x}{\int }_{1}^{0}f(y)dy.$

Let $A={\int }_{1}^{0}yf(y)dy$ and $B={\int }_{1}^{0}f(y)dy$, both are constants.

So $f(x)$ becomes,

$f(x)={e}^{x}+A+Bx{e}^{x}\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, ...(1)$

Or, $f(y)={e}^{y}+A+By{e}^{y}\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, ...(2)$

Solve, $B={\int }_{1}^{0}f(y)dy$.

$B={\int }_{1}^{0}[{e}^{y}+A+By{e}^{y}]dy,{f(y)={e}^{y}+A+By{e}^{y}}.$

$B={\int }_{1}^{0}[{e}^{y}+A+By{e}^{y}]dyf(y)={e}^{y}+A+By{e}^{y}$

Using integration by parts, $\int y{e}^{y}dy=y{e}^{y}-\int {e}^{y}={e}^{y}(y-1)$.

$\begin{gathered} B={[{e}^{y}+Ay+B{e}^{y}(y-1)]}_{1}^{0}. \ B=[{e}^{1}+A\cdot 1+0-{e}^{0}-0-B\cdot {e}^{0}(-1)]. \ B=e+A-1+B\implies A=1-e. \end{gathered}$

Finding $f(0)$ using equation (1).

$\begin{gathered} f(0)=1+A+0. \ f(0)=1+1-e,{A=1-e}. \ f(0)+e=2 \end{gathered}$

Solutions · 1