MHB How to Solve the IVP Using an Integration Factor?

[hiyum]

\[ y'=y+2te^{2t} \]

 

[topsquark]

\[ y'=y+2te^{2t} \]

This is a first order differential equation. Putting it into standard form:
[math]y' = y + 2t e^{2t}[/math]

[math]y' - y = 2t e^{2t}[/math]

Multiply both sides by [math]e^{-t}[/math] (which is the integration factor):
[math]e^{-t} y' - e^{-t} y = 2t e^t[/math].

What can you do to simplify the LHS?

-Dan

 

[Kansas Boy]

But this is NOT an "IVP" (initial value problem) because there is no initial value given.

 

[Kansas Boy]

Since this is a linear differential equation with constant coefficients we can also separate it into its 'homogeneous' and 'non-homogeneous' parts, find solutions to each, then add.

The 'homogeneous' part is y'= y which has the obvious general solution $y= Ce^t$ for any constant C.

To find a solution to the non-homogeous part try $y= (At+ B)e^{2t}$.

Then $y'= Ae^{2t}+ 2(At+ B)e^{2t}= (2At+ A+ 2B)e^{2t}$.

So we have $(2At+ A+ 2B)e^{2t}= (At+ B)e^{2t}+ 2te^{2t}$. Dividing by $e^{2t}$, $2At+ A+ 2B= At+ B+ 2t$.

This has to be true for all t. Taking t= 0, A+ 2B= B or A= -B. Taking t= 1, 2A+ A+ 2B= A+ B+ 2. 2A+ B= 2.

Since A= -B. -2B+ B= -B= 2 so B= -2 and A= 2.

$y= (2t- 2)e^{2t}$ satisfies $y'= y+ 2te^{2t}$: $y'= 2e^{2t}+ (4t- 4)e{2t}= (4t- 2)e^{2t} $ while $y+ 2te^{2t}= (2t- 2)e^{2t}+ 2te^{2t}= (4t- 2)e^{2t}$.

Therefore the general solution to $y'= y+ 2te^{2t}$ is $y= Ce^t+ (2t- 2)e^{2t}$.

 

[GJA]

I'm usually not one for replying to answered threads, especially in light of the expert help already provided by topsquark and County Boy. That said, I'm detailing another solution outline here because the method on which it's based, Duhamel's Principle, is one of my favorites. What I like about this method is its generality, including its application to inhomogeneous PDEs. Note that in the case of linear inhomogeneous ODEs, which this problem is, Duhamel's Principle is equivalent to a technique known as the Variation of Parameters.

We consider the problem
$$ y'-y = 2te^{2t}, \qquad y(0) = y_{0}\qquad (*)$$
Set $v = y-y_{0}$ and note that $v' = y'$. Then, upon substituting for $y$, $(*)$ becomes
$$v'-v = 2te^{2t} + y_{0}, \qquad v(0) = 0 \qquad(**)$$
This is an inhomogeneous linear evolution equation, whose solution is given by
$$v(t) = \int_{0}^{t}(P^{s}f)(t)ds,$$
where $(P^{s}f)(t)$ is the solution of the problem
$$v'-v = 0, \qquad v(s) = f(s) = 2se^{2s}+y_{0}.$$
See Duhamel's Principle - Wikipedia for a reference.

The solution to $v'-v = 0$ is given by $v(t) = Ce^{t}.$ The condition $v(s) = 2se^{2s}+y_{0}$ implies $C = 2se^{s}+y_{0}e^{-s}$. Thus,
$$(P^{s}f)(t) = \left[2se^{s} + y_{0}e^{-s}\right]e^{t}.$$
According to Duhamel's Principle, the solution to $(**)$ is
$$\begin{align*}
v(t) &= \int_{0}^{t}(P^{s}f)(t) ds\\
& = \int_{0}^{t}\left[2se^{s} + y_{0}e^{-s}\right]e^{t}ds\\
& = e^{t}\int_{0}^{t}2se^{s} + y_{0}e^{-s}ds
\end{align*}
$$
Using integration by parts on the $2se^{s}$ term we obtain
$$
\begin{align*}
v(t) &= e^{t}\left[2se^{s}-2e^{s} - y_{0}e^{-s}\Biggr|_{0}^{t} \right]\\
&= 2te^{2t}-2e^{2t}+(2+y_{0})e^{t} - y_{0}
\end{align*}
$$
Using $v = y - y_{0}$ to substitute back for $y$ and setting $C = 2+y_{0}$, we obtain
$$y(t) = Ce^{t} + (2t-2)e^{2t},$$
as was previously demonstrated.

 

[Kansas Boy]

That last is, I believe, equivalent to "variation of parameters". Having determined that the general solution to the associated homogeneous equation is $y= Ce^t$, we look for a solution to the entire equation of the form $y(t)= u(t)e^t$ (we allow the "parameter", C, to vary). Then $y'(t)= u'(t)e^t+ u(t)e^t$ and the equation becomes $y'- y= u'(t)e^t+ u(t)e^t- u(t)e^t= u'(t)e^t= 2te^{2t}$.

Dividing both sides by $e^t$, $u'(t)= 2te^t$.
Integrating that, $u(t)= 2te^t- e^t= (2t- 1)e^t$.
$u(t)e^t= (2t- 1)e^{2t}$ and the general solution to the equation is $y(t)= Ce^t+ (2t- 1)e^{2t}$.

 

[I like Serena]

I liked topsquark's suggestion to use the integration factor $e^{\int^t -1\cdot dt}=e^{-t}$.
Then we get:
$$y' - y = 2t e^{2t} \implies e^{-t} y' - e^{-t} y = 2t e^t \implies (e^{-t} y)' = 2t e^t \implies e^{-t}y = \int 2t e^t\,dt \implies y = e^t \int 2t e^t\,dt$$