-1.3.14 Verify the following given functions is a solution

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In summary, the given function $y=e^{{t}^2}\int_{0}^{t} e^{- s^2}\,ds+e^{t^2}$ is a solution of the differential equation $y'-2ty=1$. By using the derivative form of the Fundamental Theorem of Calculus and the product and exponential rules, we can compute $y'$ to be $y'=2te^{t^2}\int_0^t e^{-s^2}\,ds+1+2te^{t^2}$. Substituting $y$ and $y'$ into the differential equation results in the identity $1=1$, thus verifying that the given function
  • #1
karush
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$\textsf{ 1.3.14 Verify the following given functions is a solution of the differential equation}\\ \\$
$\displaystyle y'-2ty=1\\$
$\displaystyle y=e^{{t}^2}\int_{0}^{t} e^{- s^2}\,ds+e^{t^2}$
\begin{align*}\displaystyle
&=
\end{align*}

ok this one kinda baffled by the $\int$

presume to do it first before the d/dx
 
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  • #2
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?
 
  • #3
HTML:
$\displaystyle y=e^{{t}^2}\int_{0}^{t} e^{- s^2}\,ds+e^{t^2}$

thus??

$\displaystyle y'=e^{{t}^2} e^{- t^2}+e^{t^2}$
 
  • #4
karush said:
$\displaystyle y=e^{{t}^2}\int_{0}^{t} e^{- s^2}\,ds+e^{t^2}$

thus??

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

No...sorry, but you didn't differentiate anything. What you want is:

\(\displaystyle y'=2te^{t^2}\int_0^t e^{-s^2}\,ds+e^{t^2}e^{-t^2}+2te^{t^2}=2te^{t^2}\int_0^t e^{-s^2}\,ds+1+2te^{t^2}\)
 
  • #5
so why didn't the $\int$ disappear?
 
  • #6
karush said:
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. :)
 
  • #7
$\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,}\\ $
$\displaystyle y=
e^{{t}^2}\int_{0}^{t} e^{- s^2}\,ds+e^{t^2}$
$\displaystyle y'=
2te^{t^2}\int_0^t e^{-s^2}\,ds
+e^{t^2}e^{-t^2}+2te^{t^2}\\
=2te^{t^2}\int_0^t e^{-s^2}\,ds+1+2te^{t^2}$
 
  • #8
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
$1=1$
 

Related to -1.3.14 Verify the following given functions is a solution

1. What does it mean to "verify" a given function as a solution?

Verifying a given function as a solution means checking if the function satisfies the given conditions or requirements. In other words, it means determining if the function is a valid solution to the given problem.

2. What are the steps to verify a given function as a solution?

The steps to verify a given function as a solution may vary depending on the specific problem and conditions. However, in general, the steps may include plugging in the function into the original problem, simplifying the expression, and checking if it satisfies the given conditions.

3. How do you know if a given function is a valid solution?

A given function is a valid solution if it satisfies all of the given conditions or requirements. This can be determined by carefully evaluating the function and checking if it satisfies the conditions set by the problem.

4. What happens if the given function does not satisfy the conditions?

If the given function does not satisfy the conditions, then it is not a valid solution. This means that the function does not satisfy the requirements set by the problem and may need to be revised or replaced with a different function.

5. Can a given function be verified as a solution without plugging it into the original problem?

No, in order to verify a given function as a solution, it must be plugged into the original problem and evaluated. This is necessary to ensure that the function satisfies all of the given conditions and is a valid solution.

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