- #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.
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.