Need a refresher: 1st order linear diff eq

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1. Nov 13, 2015

kostoglotov

I thought I understood how to solve these sorts of equations, but apparently not..

1. The problem statement, all variables and given/known data

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 $\frac{dy}{dt} - 4y = -6e^{t}$. Using an integrating factor I get to $y = 2e^{t} + C$...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 $y = c_1 e^{t} + c_2e^{4t}$, with initial conditions $y(0) = -5$ it should become $y = 2e^{t} + 3e^{4t}$

How do I get this second solution (without using linear algebra)? Using the integrating factor method only gives me the first solution.

2. Nov 13, 2015

Staff: Mentor

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.

3. Nov 14, 2015

HallsofIvy

No, you don't. You get $$y(t)= 2e^t+ Ce^{4t}$$. You have missed the "$e^{4t}$" part.

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.

$y= c_1e^{t}+ c_2e^{4t}$ is the general solution to the second order, homogeneous, differential equation $\frac{d^2y}{dt^2}- 5\frac{dy}{dt}+ 4y= 0$.

4. Nov 15, 2015

kostoglotov

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.