if $f;R\rightarrow R$ is continuous, and $f(x)=e^{\frac{x^2}{2}}+\int_0^x tf(t)dt$, determine which of the followng is right

Question

$f(x)=e^{\frac{x^2}{2}}+\int_0^x tf(t)dt$ and the given options are

A) $5<f(\sqrt2)<6$

b) $2<f(\sqrt2)<3$

c) $3<f(\sqrt2)<4$

d) $4<f(\sqrt2)<5$

now,as it's only continuous it's surely integrable and not necessarily diffferentiable so, we have.

$f(x)=e^{\frac{x^2}{2}}+\int_0^x tf(t)dt----1$

$e^{\frac{x^2}{2}}+t \int_0^xf(t)dt-\int_0^x\int_0^xf(t)dt$

but this doesn't help at all I'm tempeted to differentiate to maybbe obtain a useful expression, so $$f'(x)=2xe^{\frac{x^2}{2}}+xf(x)$$

which allows me to use integration by parts in $1$

so $f(x)=e^{\frac{x^2}{2}}+\frac{x^2f(x)}{2}-\int_0^x \frac{x^2}{2}f'(x)dx$

which can further be simplified using $f'(x)$

which gives us $f(x)=e^{\frac{x^2}{2}}+\frac{x^2f(x)}{2}-\int_0^x \frac{x^2}{2}2xe^{\frac{x^2}{2}}+xf(x)dx$

beyond which I'm lost, as the integral seems to become zero

I'd really appreciate a HINT NOT AN EXPLICIT SOLUTON FOR NOW

Answer 1

Here's a hint that may or may not take you there as I haven't worked it out fully yet, but is a good direction to look at:

First, $f(x)$ is differentiable due to the right hand side of your given equation being differentiable. Now take your equation you got for $f'(x)$, this is a standard linear differential equation that can be solved by the method of integrating factors.

Answer 2

HINT: $$\left(e^{-\frac{x^2}{2}}f(x)\right)'=e^{-\frac{x^2}{2}}\left(f'(x)-xf(x)\right).$$

Answer 3

Hint: You already have $$f'(x)=2xe^{\frac{x^2}{2}}+xf(x)$$ and hence $$ f'(x)-xf(x)=2xe^{\frac{x^2}{2}}. \tag1$$ The integral factor of (1) is $$ \mu(x)=\exp\bigg(\int(-x)dx\bigg)=e^{-\frac{x^2}{2}} $$ and multiplying (1) by $\mu(x)$ gives $$ \bigg(e^{-\frac{x^2}{2}}f(x)\bigg)'=2x. $$ You can do the rest.