-1.3.14 Verify the following given functions is a solution

[karush]

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

ok this one kinda baffled by the $\int$

presume to do it first before the d/dx

 

[MarkFL]

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]

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

thus??

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

 

[MarkFL]

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

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]

so why didn't the $\int$ disappear?

 

[MarkFL]

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]

$\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]

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
$1=1$