Need a refresher: 1st order linear diff eq

In summary: or more precisely, second order linear differential equations that have a solution which is an infinite series.
  • #1
kostoglotov
234
6
I thought I understood how to solve these sorts of equations, but apparently not..

1. Homework Statement


In Linear Algebra I'm solving diff eqs with eigenvectors to get all the combinations that will solve for a diff eq.

The text then asked me to check my answer by going back and solving the diff eqs of the system in the usual non-linear algebra way...well, I guess I must have missed something when I learned how to solve first order linear differential equations the first time round, because something I thought was easy has stumped me.

I just need to find all the solutions to [itex]\frac{dy}{dt} - 4y = -6e^{t}[/itex]. Using an integrating factor I get to [itex]y = 2e^{t} + C[/itex]...but this is just one of the solutions. I think I remember only ever getting one answer to these sorts of questions when I did them in my calculus textbook. How do I get the second solution out of this equation.

Without initial conditions I should get [itex]y = c_1 e^{t} + c_2e^{4t}[/itex], with initial conditions [itex]y(0) = -5[/itex] it should become [itex]y = 2e^{t} + 3e^{4t}[/itex]

How do I get this second solution (without using linear algebra)? Using the integrating factor method only gives me the first solution.
 
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  • #2
kostoglotov said:
I thought I understood how to solve these sorts of equations, but apparently not..

1. Homework Statement


In Linear Algebra I'm solving diff eqs with eigenvectors to get all the combinations that will solve for a diff eq.

The text then asked me to check my answer by going back and solving the diff eqs of the system in the usual non-linear algebra way...well, I guess I must have missed something when I learned how to solve first order linear differential equations the first time round, because something I thought was easy has stumped me.

I just need to find all the solutions to [itex]\frac{dy}{dt} - 4y = -6e^{t}[/itex]. Using an integrating factor I get to [itex]y = 2e^{t} + C[/itex]...but this is just one of the solutions.

I think I remember only ever getting one answer to these sorts of questions when I did them in my calculus textbook. How do I get the second solution out of this equation.
You're not going to get a second solution. Maybe you're thinking about the solution to the homogeneous problem (y' - 4y = 0 in this case), versus the solution to the nonhomogeneous problem (y' - 4y = -6et).

The solution you got using an integrating factor is wrong on two counts: 1) it's not a solution, and 2) it shouldn't have the constant added.

The particular solution yp has the form yp = Aet. Substitute this into the nonhomogeneous equation to find A, which BTW is not equal to 2.

The solution to the homogeneous equation (also called the complementary solution) is yh = ce4t. The general solution to the nonhomogeneous problem is y = yh + yp. These two functions might be what you were thinking of when you were talking about a "second solution."

After you have found A for the particular solution, form the general solution and use the initial condition to find the constant c. This will be the solution to your initial value problem.
kostoglotov said:
Without initial conditions I should get [itex]y = c_1 e^{t} + c_2e^{4t}[/itex], with initial conditions [itex]y(0) = -5[/itex] it should become [itex]y = 2e^{t} + 3e^{4t}[/itex]

How do I get this second solution (without using linear algebra)? Using the integrating factor method only gives me the first solution.
 
  • #3
kostoglotov said:
I thought I understood how to solve these sorts of equations, but apparently not..

1. Homework Statement


In Linear Algebra I'm solving diff eqs with eigenvectors to get all the combinations that will solve for a diff eq.

The text then asked me to check my answer by going back and solving the diff eqs of the system in the usual non-linear algebra way...well, I guess I must have missed something when I learned how to solve first order linear differential equations the first time round, because something I thought was easy has stumped me.

I just need to find all the solutions to [itex]\frac{dy}{dt} - 4y = -6e^{t}[/itex]. Using an integrating factor I get to [itex]y = 2e^{t} + C[/itex]
No, you don't. You get [tex]y(t)= 2e^t+ Ce^{4t}[/tex]. You have missed the "[itex]e^{4t}[/itex]" part.

...but this is just one of the solutions. I think I remember only ever getting one answer to these sorts of questions when I did them in my calculus textbook. How do I get the second solution out of this equation.
Actually, you have an infinite number of solutions, a different one for every different value of C. I think you mean that there is only one independent solution. But that is to be expected for first order differential equations. It is second order differential equations that have two different independent solutions. (In general, n-th order linear differential equations have n independent solutions.

Without initial conditions I should get [itex]y = c_1 e^{t} + c_2e^{4t}[/itex], with initial conditions [itex]y(0) = -5[/itex] it should become [itex]y = 2e^{t} + 3e^{4t}[/itex]

How do I get this second solution (without using linear algebra)? Using the integrating factor method only gives me the first solution.
[itex]y= c_1e^{t}+ c_2e^{4t}[/itex] is the general solution to the second order, homogeneous, differential equation [itex]\frac{d^2y}{dt^2}- 5\frac{dy}{dt}+ 4y= 0[/itex].
 
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  • #4
HallsofIvy said:
No, you don't. You get [tex]y(t)= 2e^t+ Ce^{4t}[/tex]. You have missed the "[itex]e^{4t}[/itex]" part.

Yep, I was so used to treating that C as something that you can subsume other constants into, and still call it C, that I didn't realize I was trying to subsume a function of t into it :) That constant of integration is such a common blind-spot.
 

Related to Need a refresher: 1st order linear diff eq

1. What is a first-order linear differential equation?

A first-order linear differential equation is a mathematical equation that describes the relationship between a function and its derivative, where the highest power of the derivative is 1. It can be written in the form of dy/dx + P(x)y = Q(x), where P(x) and Q(x) are functions of x.

2. What is the general solution to a first-order linear differential equation?

The general solution to a first-order linear differential equation is a function that satisfies the equation for all possible values of the independent variable. It can be found by solving the equation using various methods, such as separation of variables, integrating factor, or variation of parameters.

3. How is a first-order linear differential equation different from other types of differential equations?

A first-order linear differential equation is different from other types of differential equations because it only involves the first derivative of the function. Other types of differential equations, such as second-order or higher-order, involve higher powers of the derivative. This makes solving first-order linear differential equations more straightforward and often requires less advanced methods.

4. Can a first-order linear differential equation have multiple solutions?

Yes, a first-order linear differential equation can have multiple solutions. This is because the general solution to the equation includes an arbitrary constant, which can take on different values. However, any particular solution to the equation must satisfy any given initial conditions.

5. What are some real-world applications of first-order linear differential equations?

First-order linear differential equations have many applications in science and engineering, such as modeling population growth, predicting the decay of radioactive substances, analyzing electrical circuits, and describing the motion of objects subject to a constant force. They are also used in economics and finance to model growth and decay in various systems.

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