2.1.2 Find the general solution of y'−2y=t^{2}e^{2t}

In summary, the general solution of $y'-2y=t^{2}e^{2t}$ is $y=e^{2t}\left(\dfrac{t^3}{3} + C\right)$ and as $t \to \infty$, the solution behaves like $y \to \infty$ due to the exponential growth of $e^{2t}$.
  • #1
karush
Gold Member
MHB
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Find the general solution of
$y'−2y=t^{2}e^{2t}$
and use it to determine how solutions behave as
$t \to \infty$

ok presume the first thing to do is to find $u{x}$
$\exp{\displaystyle\int{2} y}=e^{-2} or \dfrac{1}{e^2}$
 
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  • #2
is the DE
$y' - 2y = t^2e^{2t}$ ?

if so, then the integrating factor is $\mu = e^{\int -2 \, dt} = e^{-2t}$

$e^{-2t} \cdot y' + (-2e^{-2t} \cdot y) = t^2$


$(e^{-2t} \cdot y)' = t^2$

$e^{-2t} \cdot y = \dfrac{t^3}{3} + C$

$y = e^{2t}\left(\dfrac{t^3}{3} + C\right)$
 
  • #3
do you mean

\(\displaystyle y^\prime - 2y = t^2 e^{2t}\)

if so we have

\(\displaystyle y^\prime + p(t) = q(t)\\
p(t)=-2,~q(t) = t^2e^{2t}\)

The integrating factor is given by

\(\displaystyle \nu(t) = e^{\int p(t)~dt} = e^{-2t}\)

\(\displaystyle e^{-2t}y^\prime - 2 y e^{-2t} = t^2\\

\dfrac{d}{dt}\left(e^{-2t}y\right) = t^2 \\

e^{-2t} y = \dfrac 1 3 t^3 + C\\

y = e^{2t}\left(\dfrac 1 3 t^3 + C\right)

\)
 
  • #4
y- 2y? That's equal to -y! Do you mean $y'- 2y= t^2e^{2t}$?
You say "the first thing to do is find ux". That makes no sense because there is no "u" or "x" in the problem!

I presume you mean you are looking for an "integrating factor". That would be a function, u(t), such that u(t)y'- 2u(t)y= (u(t)y)'. Since (u(t)y)'= u(t)y'+ u'(t)y we must have u'(t)y= -2u(t)y so that u'(t)= -2u(t). Then u'(t)/u(t)= -2. Integrating, ln(u)= -2t+ C and $u(t)= C'e^{-2t}$ where $C'= e^C$.

Multiplying both sides of $y'- 2y= t^2e^{2t}$ by $e^{-2t}$ (since we can use any integrating factor we can take C'= 1) we get $e^{-2t}y'- 2e^{-2t}y= (e^{-2t}y)'= t^2$.

Integrating both sides $e^{-2t}y= \frac{1}{3}t^3+ C$ so that

$y(t)= Ce^{2t}+ \frac{1}{3}t^3e^{2t}$.
 
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  • #5
romsek said:
do you mean

\(\displaystyle y^\prime - 2y = t^2 e^{2t}\)

if so we have

\(\displaystyle y^\prime + p(t) = q(t)\\
p(t)=-2,~q(t) = t^2e^{2t}\)

The integrating factor is given by

\(\displaystyle \nu(t) = e^{\int p(t)~dt} = e^{-2t}\)

\(\displaystyle e^{-2t}y^\prime - 2 y e^{-2t} = t^2\\

\dfrac{d}{dt}\left(e^{-2t}y\right) = t^2 \\

e^{-2t} y = \dfrac 1 3 t^3 + C\\

y = e^{2t}\left(\dfrac 1 3 t^3 + C\right)

\)
yes my bad #2

Screenshot 2021-08-20 2.34.38 PM.png
 

FAQ: 2.1.2 Find the general solution of y'−2y=t^{2}e^{2t}

1. What is the equation for finding the general solution of y'−2y=t^{2}e^{2t}?

The equation for finding the general solution of y'−2y=t^{2}e^{2t} is y = Ce^{2t} + \frac{1}{4}t^{2}e^{2t} - \frac{1}{2}te^{2t} + \frac{1}{4}e^{2t}, where C is a constant.

2. How do you solve for the general solution of y'−2y=t^{2}e^{2t}?

To solve for the general solution of y'−2y=t^{2}e^{2t}, you can use the method of variation of parameters or the method of undetermined coefficients. Both methods involve finding the complementary solution and particular solution, and then adding them together to get the general solution.

3. Can you explain the concept of variation of parameters in relation to y'−2y=t^{2}e^{2t}?

Variation of parameters is a method used to solve non-homogeneous linear differential equations, such as y'−2y=t^{2}e^{2t}. It involves assuming a general solution in the form of y = u(t)e^{2t}, where u(t) is a function to be determined. This allows us to find the particular solution and then add it to the complementary solution to get the general solution.

4. What is the particular solution of y'−2y=t^{2}e^{2t}?

The particular solution of y'−2y=t^{2}e^{2t} is \frac{1}{4}t^{2}e^{2t} - \frac{1}{2}te^{2t} + \frac{1}{4}e^{2t}.

5. How do you know if you have found the correct general solution for y'−2y=t^{2}e^{2t}?

To check if you have found the correct general solution for y'−2y=t^{2}e^{2t}, you can substitute the solution into the original equation and see if it satisfies the equation. You can also take the derivative of the solution and see if it matches the original equation.

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