Second Linear ODE

Question

Confused on how I can approach this ODE, and I just need some help on figuring out what to do

$y''-4y'-21=5e ^{-t}$ with initial conditions $y(0)=0$ and $y'(0)=4$

After Laplace transforming, I got: $(\frac{(5)}{(s+1)}{+ 4+\frac{(21)}{(s+1)}}) /( s^2-4s) $

And now I am stuck on what to do here

Answer 1

I don't know about Laplace transformation but I would solve it a little differently. Look at it the following way: $$y''-4y' = 5e^{-t} + 21$$ Now solving the homogenous equation $y''-4y'=0$ is not too complicated. It has the characteristic polynomial $t^2-4t$ with zeros at $t=0$ and $t=4$. Therefore the solution to the homogenous part equation is $c_1 e^{4t} + c_2e^{0t} = c_1e^{4t} + c_2$.

Now we need to find a special solution to the inhomogenous problem. To be honest I don't know how to do that systematically, but a little "good guessing" can get us there: How do we get the $+21$? Let us put a $\frac{21}{4} t$ in the solution, so that the $4y'$ produces our $21$. How do we get $5e^{-t}$? Because of the fact that the second derivative of $e^{-t}$ is $e^{-t}$ and the first derivative is $-e^{-t}$ we can find the special solution $$u_{sp} = e^{-t} - \frac{21}{4} t$$

Then the whole solution to this equation is: $$u= c_1e^{4t}+c_2 + e^{-t} - \frac{21}{4} t$$

Now we try to find $c_1, c_2$ via the initial conditions. Note $0=u(0) = c_1 + c_2+1$ and $4=u'(0) = 4c_1 -1- \frac{21}{4}$ Solving this we get $c_1 = \frac{41}{16}$ and $c_2 = \frac{-57}{16}$.

If anybody knows how we could have gotten the special solution in a more systematic way feel free to let me know. If the Laplace transform method would be easier than my approach to solving linear ODEs with constant coefficients I am also happy if you let me know because I don't know how to do Laplace transformation.

Answer 2

Because you didn't write down your process i'll solve the whole thing here.

Let $y''-4y'-21=5e^{-t}:=f(y(t))=f(y)$

1. Take the Laplace Transform of each term.
2. Plug in initial value(s) of $y^{(i)}(t_0)$ into the equation.
3. Solve for $Y(s)$.
4. Take the Inverse Laplace Transform.

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1. $$\mathscr{L}\{f(y)\}=\mathscr{L}\{y''-4y'-21=5e^{-t}\}=[s^2Y(s)-sy(0)-y'(0)]-4[sY(s)-y(0)]-\dfrac{21}{s}=5\dfrac{1}{s+1}$$

You can only arrive to $\frac{1}{s-a}$ if you have a $e^{at}$ function.

2. $$\mathscr{L}\{f(y,y(0),y'(0))\}=[s^2Y(s)-0-4]-4[sY(s)-0]-\dfrac{21}{s}=\dfrac{5}{s+1}$$ $$s^2Y(s)-4-4sY(s)-\dfrac{21}{s}=\dfrac{5}{s+1}$$

3. $$s^2Y(s)-4sY(s)=\dfrac{5}{s+1}+\dfrac{21}{s}+4$$ $$Y(s)(s^2-4s)=\dfrac{5}{s+1}+\dfrac{21}{s}+4$$ $$Y(s)=\dfrac{5}{(s+1)(s^2-4s)}+\dfrac{21}{s(s^2-4s)}+\dfrac{4}{(s^2-4s)}$$

4. $$Y(s)=\dfrac{5}{s(s+1)(s-4)}+\dfrac{21}{s^2(s-4)}+\dfrac{4}{s(s-4)}$$

$$\mathscr{L}^{-1}\{Y(s)\}=y(t)=-\dfrac{5}{4}e^{0t} + e^{-t} + \dfrac{1}{4}e^{4t} - e^{0t} + e^{4t} -\dfrac{21}{16}-\dfrac{21}{4}t+\dfrac{21}{16}e^{4t}$$

$$y(t)=-\dfrac{5}{4} + e^{-t} + \dfrac{1}{4}e^{4t} - 1 + e^{4t} -\dfrac{21}{16}-\dfrac{21}{4}t+\dfrac{21}{16}e^{4t}$$

$$y(t)=-\dfrac{20}{16} + e^{-t} + \dfrac{4}{16}e^{4t} - \dfrac{16}{16} + \dfrac{16}{16}e^{4t} -\dfrac{21}{16}-\dfrac{21}{4}t+\dfrac{21}{16}e^{4t}$$

$$\boxed{y(t)=\dfrac{41}{16}e^{4t}-\dfrac{57}{16}+e^{-t}-\dfrac{21}{4}t}$$

Which checks out by @Lukas' answer.

Answer 3

@LordCommander did it the way that was asked, but it is actually a really simple variable substitution problem. Original problem:

$$ y'' - 4y' = 5e^{-t} + 21 $$

Substitute $w = y'$:

$$ w' - 4w = 5e^{-t} + 21 \\ dw - 4w\, dt = 5e^{-t}\,dt + 21\,dt $$

Multiply by an integrating factor of $e^{-4t}$:

$$ e^{-4t}\, dw - 4 e^{-4t}\, dt = 5e^{-5t}\, dt + 21e^{-4t}\, dt \\ \int e^{-4t}\, dw - 4 e^{-4t}\, dt = \int 5e^{-5t}\, dt + 21e^{-4t}\, dt \\ w\,e^{-4t} = e^{-5t} - (1/4) 21 e^{-4t} + C \\ w = e^{-t} - (21/4) + Ce^4t $$

Now we can substitute $y' = w$ back in and solve again.

$$ y' = e^{-t} - (21/4) + Ce^4t \\ dy = e^{-t}\,dt - (21/4)\,dt + Ce^4t\,dt \\ \int dy = \int e^{-t}\,dt - (21/4)\,dt + Ce^4t\,dt \\ y = e^{-t} - (21/4)t + Ce^4t + D $$