First order differential equations

[DmytriE]

Homework Statement

Consider the first order differential equation

\frac{dx(t)}{dt} + ax(t) = f(t), x(0) = x_{0}, t\geq0​

Suppose the "input signal" f(t)=e^{-t}, t\geq0. (a) Find the solution to the equation. Find a condition on the parameter a so that the solution of the (forced) system approaches zero as t→∞.

Homework Equations

\frac{dy}{dt} + p(x)y = 0

The Attempt at a Solution

Setup as a homogenous equation therefore f(t) = 0.
\frac{dx(t)}{dt} + ax(t) = 0

\frac{dx(t)}{x(t)} = -a*dt

ln(x(t)) = -at

x(t) = e^{-at}

I don't know how to proceed any further...

 

[BvU]

What you found satisfies the homogeneous equation. You are still missing an integration constant.
Now you need one particular solution that satisfies the inhomeogeneous equation.
A times homogeneous solution + particular solution is ' the ' general solution

 

[DmytriE]

Is the integration constant e^{\int x(t) dt} or e^{\int a dt}?

I guess the only integration factor that makes sense is the later of the two.

 

[BvU]

No, I mean x(t) = Ce^{-at} also satisfies the homogeneous equation.

 

[BvU]

Next thing to try: C = C(t). Put this in the inhomogeneous equation and see what equation you get for C.

 

[DmytriE]

I'm sorry but you've now lost me. Why am I trying C=C(t)?

 

[BvU]

Well, the e^{-at} part of x(t) = C(t)e^{-at} helps you get rid of the +a x(t) term. In return, you are left with a differential equation for C(t) that you may be able to solve.

 

[pasmith]

Homework Statement

Consider the first order differential equation

\frac{dx(t)}{dt} + ax(t) = f(t), x(0) = x_{0}, t\geq0​

Suppose the "input signal" f(t)=e^{-t}, t\geq0. (a) Find the solution to the equation. Find a condition on the parameter a so that the solution of the (forced) system approaches zero as t→∞.

Homework Equations

\frac{dy}{dt} + p(x)y = 0

The Attempt at a Solution

Setup as a homogenous equation therefore f(t) = 0.
\frac{dx(t)}{dt} + ax(t) = 0

\frac{dx(t)}{x(t)} = -a*dt

ln(x(t)) = -at

x(t) = e^{-at}

I don't know how to proceed any further...

Either use Method of undetermined coefficients or use the integrating factor e^{at}.

 

[Chestermiller]

Either use Method of undetermined coefficients or use the integrating factor e^{at}.

Yes. This screams out for one to use the integrating factor.

 

[DmytriE]

How can you tell when a first differential needs to use an integrating factor?

 

[Saitama]

How can you tell when a first differential needs to use an integrating factor?

The differential equation you have presented is a linear differential equation of first order and this immediately tells us to use the integrating factor.

 

[Dick]

How can you tell when a first differential needs to use an integrating factor?

In this case, you don't need an integrating factor. Just put $x(t)=Ce^{-t}$ into the equation and solve for C. You can't always do that but it's a nice shortcut here. Can you see why you could guess it would work? In general, guessing a trial solution is often a good option. When you can.

 

[Chestermiller]

How can you tell when a first differential needs to use an integrating factor?

Look at the left side of your equation. "a" is a constant, so it is certainly going to integrate nicely. If "a" can be integrated easily with respect to t, that's your first indicator. Now look at the right side of the equation. It's a function only of t. So, when you multiply both sides by the integrating factor, the right side will still only be a function only of t (that can be integrated). That's your second indicator.

Chet

 

[vanhees71]

Perhaps it's good to give a general derivation of the integrating factor for 1st-order linear ODEs. The standard form of such an ODE is
y'(x)+p(x) y(x)=q(x),
where p and q are given functions and the equation is to besolved for y(x). We look for the general solution.

The equation would be much easier, if we had a total derivative on the left-hand side, but that's not too difficult to achieve. You just substitute
z(x)=v(x) y(x).
From the product rule you get
z'(x)=v(x) y'(x)+v'(x) y(x).
Multiplying the ODE with v(x) gives
v(x) y'(x)+p(x) v(x) y(x)=q(x) v(x).
To make the left-hand side equal to z'(x) you obviously have to determine v(x) so that
v'(x)=p(x) v(x)
This is very easily solved, because you can write
\frac{\mathrm{d}}{\mathrm{d} x} \ln \left (\frac{v(x)}{v_0} \right )=p(x).
This gives
v(x)=v_0 \exp \left (\int_{0}^{x} \mathrm{d} x' p(x') \right ).
Here v_0 \neq 0 is an arbitrary integration constant.

Thus, choosing v(x) in this way we can write our ODE as
z'(x)=q(x) v(x)
with the general solution
z(x)=z_0 + \int_0^{x} \mathrm{d} x' q(x') v(x').
Finally we have
y(x)=\frac{z(x)}{v(x)}=\frac{z_0}{v(x)}+\frac{1}{v(x)} \int_0^{x} \mathrm{d} x' q(x') v(x').
As you see, the choice of v_0 doesn't affect your solution in any way, because it only appears as an overall factor in front of the integration constant z_0, and this can be lumped into the overall integration constant of the general solution of the homogeneous equation. Finally we have the general solution of the ODE
y(x)=\exp \left (-\int_0^{x} \mathrm{d} x' p(x') \right ) \left [y_0+ \int_0^x \mathrm{d} x' q(x') \exp \left ( \int_0^x \mathrm{d} x'' p(x'') \right ) \right].
Here y_0 is given by the initial-value condition y(0)=y_0.

 

[DmytriE]

Thank you everyone! I finally figured it out using integrating factors. As for guessing, would you guess Ce^-t as a good solution because the function is forced by e^-t?

 

[Dick]

Thank you everyone! I finally figured it out using integrating factors. As for guessing, would you guess Ce^-t as a good solution because the function is forced by e^-t?

It's a good guess because of the other side of the equation, if you differentiate $e^{-t}$ or multiply it by $a$, you just get back multiples of $e^{-t}$.