-2.1.1 DE Find the general solution

In summary, the notation "-2.1.1 DE" refers to a differential equation that involves a function and its derivatives. "Find the general solution" means to find an expression that satisfies the equation for all possible values of the independent variable. To solve this type of differential equation, various mathematical techniques can be used such as separation of variables, integrating factors, or the method of undetermined coefficients. The difference between a particular solution and a general solution is that a particular solution is for specific values of the independent variable, while a general solution is for all possible values. It is important to find the general solution as it allows us to find all possible solutions and understand the behavior of the system being studied. A differential equation can have multiple solutions
  • #1
karush
Gold Member
MHB
3,269
5
[a] Find the general solution of $y^\prime + 3y=t+e^{-2x}\quad \dfrac{dy}{dx}+f(x)y=g(x)$
\$\begin{array}{lll}
\textsf{Similarly} & \dfrac{dy}{dx}+Py=Q\\
\textsf{hence} & \mu(x)=\exp\left(\int f(x)\,dx\right)\\
\textsf{then} & \mu^\prime(x)=\exp\left(\int f(x)\,dx\right)f(x) \\
\textsf{then} & \mu(x)+y'=\mu(x)g(x)\\
\textsf{integrating factor} & \mu(x)=e^{3x}\\
\textsf{multiplying} & e^{3x}y'+3e^{3x}y=xe^{3x}+e^{x}\\
\textsf{rewriting the LHS} & \dfrac{d}{dx}\left(e^{3x}y\right)=xe^{3x}+e^{x}\\
\end{array}$
determine how the solution behave as $t \to \infty$
$ce^{-3x}+\dfrac{x}{3}-\dfrac{1}{9}+e^{-2x}$
y is asymptotic to $\dfrac{t}{3} −\dfrac{1}{9} \textit{ as } t \to \infty$

ok i think this is correct just could be worded better
maybe some typos
suggestions, complaints, or ?
ssct.png
 
Last edited:
Physics news on Phys.org
  • #2
I’m assuming the independent variable is $t$ …

$y’ +3y = t + e^{-2t}$

$\mu(t) = e^{3t}$

$y’ \cdot e^{3t} + y \cdot 3e^{3t} = t \cdot e^{3t} + e^t$

$(y \cdot e^{3t})’ = t\cdot e^{3t} + e^t$

$y \cdot e^{3t} = \dfrac{t}{3} \cdot e^{3t} - \dfrac{1}{9} \cdot e^{3t} + e^t + C$

$y = \dfrac{3t-1}{9} + e^{-2t} + Ce^{-3t}$

$y$ is asymptotic to the line you cited.
 
  • #3
here is where it came from #1
2.1.png

asymptotic ??
 
  • #4
karush said:
here is where it came from #1
View attachment 11299
asymptotic ??

$y=\dfrac{3t-1}{9}$ is a slant asymptote.

as $t \to \infty$, the value of $y$ is very close to the value of the line $y=\dfrac{3t-1}{9}$ and it gets very close rather quickly because the two exponential terms approach zero very quickly.
 
  • #5
$y'+ 3y= t+ e^{-2t}$

(The original post has $y'+ 3y= t+ e^{-2x}$ but that just doen't make sense! The independent variable has to be either x or t. I have chosen t.)

Another method:
The "associated homogenous equation" is y'+ 3y= 0.
$\frac{dy}{dt}= -3y$
$\frac{dy}{y}= -3dt$
$ln(y)=- 3t+ C$
$y= e^{-3t+ C}= e^Ce^{-3t}= C'e^{-3t}$
where $C'= e^C$

Since that does not involve either t or $e^{-2t}$ we try a solution to the entire equation of the form $y= At+ B+ Ce^{-2t}$.
(The "method of undetermined coefficients")

$y'+ 3y= A- 2Ce^{-2t}+ 3At+ 3B+3Ce^{-2t}=3At+ (A+ 3B)+ Ce^{-2t}= t+ e^{-2t}$

So 3A= 1, A+ 3B= 0, and C= 1. A= 1/3, and B= -1/9.
$y(t)= C'e^{-3t}+ t/3- 1/9+ e^{-2t}$
 
  • #6
$y'+ 3y= t+ e^{-2t}$

Yet another method (variation of parameters):
Calculating as before that the general solution to the associated homogenous equation is $C'e^{-3t}$ we seek a solution to the entire equation of the form $y= u(x)e^{-3t}$ where u(x) is a function to be determined.
(We are allowing the "parameter", C', to "vary".)

$y'+ y= u'(x)e^{-3t}- 3u(x)e^{-3t}+ 3u(x)e^{-3t}= u'(x)e^{-3t}= t+ e^{-2t}$
$u'(t)= te^{3t}+ e^t$

And now just integrate. The integral of $e^t$ is of course $e^t$. To integrate $te^{3t}$ use "integration by parts", taking u= t and $dv= e^{3t}dt$. Then du= dt and $v= \frac{1}{3}e^{3t}$ so
$\int te^{3t}dt= \frac{1}{3}te^{3t}- \frac{1}{3}\int e^{3t}dt= \frac{1}{3}te^{3t}- \frac{1}{9}e^{3t}$.
$u(t)= \frac{1}{3}te^{3t}- \frac{1}{9}e^{3t}+ e^t$.

Then $u(t)e^{-3t}= \frac{1}{3}t- \frac{1}{9}+ e^{-2t}$
and
$y(t)= C'e^{-3t}+ \frac{1}{3}t- \frac{1}{9}+ e^{-2t}$.
 

FAQ: -2.1.1 DE Find the general solution

1. What is the meaning of "-2.1.1 DE Find the general solution"?

The notation "-2.1.1 DE Find the general solution" is commonly used in mathematics and science to represent a problem or equation that requires finding a general solution. In this case, the "DE" stands for "differential equation", indicating that the problem involves finding a solution to a differential equation. The "-2.1.1" refers to the specific problem number or section in which the general solution can be found.

2. How do you find the general solution to a differential equation?

To find the general solution to a differential equation, you must first solve the equation by finding the antiderivative of the given function. This will result in a family of solutions, as the general solution will include a constant of integration. This constant can then be determined by using initial conditions or boundary conditions specific to the problem.

3. What is the difference between a general solution and a particular solution?

A general solution is a solution that includes a constant of integration and can be applied to a wide range of problems. On the other hand, a particular solution is a specific solution that is obtained by assigning specific values to the constants in the general solution. In other words, a particular solution satisfies both the differential equation and any given initial or boundary conditions.

4. Why is it important to find the general solution to a differential equation?

Finding the general solution to a differential equation is important because it allows for a wide range of problems to be solved using a single solution. This saves time and effort as it eliminates the need to solve the same equation multiple times for different initial or boundary conditions. Additionally, the general solution provides a better understanding of the behavior and properties of the equation.

5. Can the general solution to a differential equation be expressed in different forms?

Yes, the general solution to a differential equation can be expressed in different forms. This is because the constant of integration can take on different values, resulting in different solutions. However, all forms of the general solution are equivalent and can be transformed into one another using algebraic manipulation.

Similar threads

Replies
5
Views
1K
Replies
1
Views
1K
Replies
1
Views
1K
Replies
4
Views
3K
Replies
4
Views
1K
Replies
4
Views
1K
Replies
3
Views
1K
Replies
4
Views
1K
Back
Top