# [SOLVED]1.3.14 Verify the following given functions is a solution

#### karush

##### Well-known member
$\textsf{ 1.3.14 Verify the following given functions is a solution of the differential equation}\\ \\$
$\displaystyle y'-2ty=1\\$
$\displaystyle y=e^{{t}^2}\int_{0}^{t} e^{- s^2}\,ds+e^{t^2}$
\begin{align*}\displaystyle
&=
\end{align*}

ok this one kinda baffled by the $\int$

presume to do it first before the d/dx

#### MarkFL

##### Administrator
Staff member
You are given $y$, and from this you need to compute $y'$, so you can then substitute both into the given ODE and see if an identity results. Also, recall the derivative form of the FTOC:

$$\displaystyle \frac{d}{dx}\left(\int_a^x f(t)\,dt\right)=f(x)$$

So, use this along with the product and exponential rules to compute $y'$...what do you get?

#### karush

##### Well-known member
HTML:
$\displaystyle y=e^{{t}^2}\int_{0}^{t} e^{- s^2}\,ds+e^{t^2}$

thus??

$\displaystyle y'=e^{{t}^2} e^{- t^2}+e^{t^2}$

#### MarkFL

##### Administrator
Staff member
$\displaystyle y=e^{{t}^2}\int_{0}^{t} e^{- s^2}\,ds+e^{t^2}$

thus??

$\displaystyle y'=e^{{t}^2} e^{- s^2}+e^{t^2}$
No...sorry, but you didn't differentiate anything. What you want is:

$$\displaystyle y'=2te^{t^2}\int_0^t e^{-s^2}\,ds+e^{t^2}e^{-t^2}+2te^{t^2}=2te^{t^2}\int_0^t e^{-s^2}\,ds+1+2te^{t^2}$$

#### karush

##### Well-known member
so why didn't the $\int$ disappear?

#### MarkFL

##### Administrator
Staff member
so why didn't the $\int$ disappear?
Because when we apply the product rule, it doesn't get differentiated in one of the resulting terms. #### karush

##### Well-known member
$\textit{so...,}\\$
$(uv)'+u'=u'v+uv'+u'$
\textit{thus,}\\ \begin{align*}\displaystyle {u}&={e^{{t}^2}} &u'&=2te^{t^2} \\ {v}&={\int_{0}^{t} e^{- s^2}\,ds} &v'&=e^{-t^2} \end{align*}
$\textit{as given,}\\$
$\displaystyle y= e^{{t}^2}\int_{0}^{t} e^{- s^2}\,ds+e^{t^2}$
$\displaystyle y'= 2te^{t^2}\int_0^t e^{-s^2}\,ds +e^{t^2}e^{-t^2}+2te^{t^2}\\ =2te^{t^2}\int_0^t e^{-s^2}\,ds+1+2te^{t^2}$

#### karush

##### Well-known member
 then $\displaystyle y'-2ty=1\\$

$2te^{t^2}\int_0^t e^{-s^2}\,ds+1+2te^{t^2} -2t(e^{{t}^2}\int_{0}^{t} e^{- s^2}\,ds+e^{t^2})=1$
by distribution and canceling
$1=1$