How to Solve the IVP Using an Integration Factor?

• MHB
• hiyum
In summary, the solution to the equation is $y= Ce^t+ (2t- 1)e^{2t}$, where $e^{-t} y' - e^{-t} y = 2t e^t$.
hiyum
$y'=y+2te^{2t}$

hiyum said:
$y'=y+2te^{2t}$
This is a first order differential equation. Putting it into standard form:
$$\displaystyle y' = y + 2t e^{2t}$$

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

Multiply both sides by $$\displaystyle e^{-t}$$ (which is the integration factor):
$$\displaystyle e^{-t} y' - e^{-t} y = 2t e^t$$.

What can you do to simplify the LHS?

-Dan

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

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 $y= Ce^t+ (2t- 2)e^{2t}$.

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.

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

1. What is an IVP?

An IVP (initial value problem) is a type of mathematical problem that involves finding a function or equation that satisfies certain conditions at a specific starting point, or initial value.

2. How do you find the solution y to an IVP?

To find the solution y to an IVP, you need to use a mathematical method or technique, such as separation of variables, substitution, or variation of parameters. These methods involve manipulating the given equation and initial conditions to solve for the unknown function y.

3. What is the importance of finding the solution y to an IVP?

Finding the solution y to an IVP is important because it helps us understand and model real-world phenomena, such as population growth, chemical reactions, and electrical circuits. It also allows us to make predictions and analyze the behavior of systems over time.

4. Are there different types of IVPs?

Yes, there are different types of IVPs, such as first-order and second-order problems, as well as systems of differential equations. The methods used to solve these problems may vary, but the basic concept of finding a function that satisfies certain conditions remains the same.

5. Can technology be used to find the solution y to an IVP?

Yes, technology such as calculators and computer software can be used to find the solution y to an IVP. These tools can help with calculations and graphing, making it easier to visualize and understand the solution. However, it is still important to understand the mathematical concepts and methods behind finding the solution.

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