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$dy/dt$ = 0.8y with y(0) = -0.8 , why does the result of this question end up being so different from the formula?

Mathematics Asked by GORO1 on January 23, 2021

So I’m given a dynamic system $dy/dt$ = 0.8t with y$_0$ = -0.8.

So the Theorem in the book I’m studying says $dx/dt$ = kx
With initial value x$_0$ and k being an ordinary constant.

x is just an arbitrary variable, surely I can bring this principle to ‘y’ as well, right?

The solution is then $e^{kt}$$x_0$

The understanding is that x will grow or decay exponentially.

The answer I originally got with this understanding was ${-0.8}$$e^{0.8t}$.
Seems simple enough, right? Except that was not the answer. I went online and found the answer, but I do not understand how or why we come to that conclusion and ditch the theorem.

The answer I found was actually $0.4t^2$+C

C being -0.8.
And the final answer being $0.4t^2$ – 0.8.

Why do we end up doing this instead of the theorem I brought up earlier? And with this being used as a tool obviously being Integration. When do I draw the line between using an Integration or an Exponential?

Can I use Integration instead of Exponentials?

One Answer

You need to be careful with the variables and functions.

The differential equation you which to solve is given by

$$frac{dy(t)}{dt} = 0.8 y(t)$$

and has solution given by (with $y_0 = y(0) = -0.8$)

$$y(t) = -0.8 e^{0.8t}$$

The solution you propose is, however (ignoring initial conditions)

$$y(t) = 0.4t^2 + C$$

Your solution makes sense if the differential equation is

$$frac{dy(t)}{dt} = 0.8 t$$

You need to be very careful about the difference between the two.

Correct answer by Eevee Trainer on January 23, 2021

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