General solution to differential equations

In summary: So if you have a particular solution to a non-homogeneous DE, then you can use the linear algebra methods to find all the other solutions that differ from that particular solution by a vector in an n dimensional vector space.
  • #1
dionysian
53
1
I’m reviewing differential equations after taking the course about 5-6 years ago and I have a couple of questions about the solutions of differential equations.
1) First why is the general form of the solution to linear homogenous differential equations, with non-equal and real roots to the complementary equation, of the forum
[tex]y = {e^{{\alpha _1}x}} + {e^{{\alpha _2}x}} + \cdot \cdot \cdot + {e^{{\alpha _n}x}}[/tex]
And real and equal roots to the complementary equation
[tex]y = {e^{{\alpha _1}x}} + x{e^{{\alpha _2}x}} + \cdot \cdot \cdot + {x^{n - 1}}{e^{{\alpha _n}x}}[/tex]

It seems that the resources on this subjects simply tell you this is the form of the solution with very little explanation of why this is known to be the form of the solution. If anyone can give me some insight into this I would appreciate it. Thanks

2) The second question is how do we know that the general solution to the nth order non-homogenous differential equation is: [tex]y(t) = {y_p}(t) + {y_h}(t)[/tex]
 
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  • #2
1) derive it - you'll see why.

2) ... a particular solution to the non-homogeneous and the general solution of the homogeneous part? Starting from 1 and because any superposition of solutions is itself a solution. It's like the homogeneous solution provides a set of lines and the particular solution selects one of them.

It is difficult to answer why questions in math. All the manipulations are exercises in tautology so the thing is because the thing is. The basic method to finding solutions to DEs is to guess at them ... what you are learning is rules of thumb for the guesswork. For instance, a homogeneous DE means that the derivative is proportional to the function ... and the only function that has that property is the exponential or functions derived from exponentials like trig functions, so you guess exponentials. The rest follows from there.
 
  • #3
Have you taken linear algebra? The basic theory is that the set of all solutions to an nth order linear homogeneous differential equation forms an n dimensional vector space. And one can show that, given any single solution, f(x), to a non-homogeneous linear differential equation then any solution minus f(x) lies in an n dimensional vector space.
 

What is a general solution to a differential equation?

A general solution to a differential equation is a function that satisfies the given differential equation. It includes all possible solutions to the equation, and can be used to find specific solutions by incorporating initial conditions.

How is a general solution different from a particular solution?

A particular solution is a specific solution to a differential equation, while a general solution includes all possible solutions. A particular solution can be found by substituting specific values for the arbitrary constants in the general solution.

What is the process for finding a general solution to a differential equation?

The process for finding a general solution to a differential equation involves first solving the equation using integration or other techniques, then incorporating any initial conditions to find specific values for the arbitrary constants. This results in a general solution that can be used to find particular solutions.

Can a differential equation have multiple general solutions?

Yes, a differential equation can have multiple general solutions. This is because the arbitrary constants in the general solution can have different values, resulting in different functions that satisfy the equation.

How can a general solution be verified?

A general solution can be verified by substituting it into the original differential equation and checking that it satisfies the equation. Additionally, it can be checked by using initial conditions to find a particular solution and confirming that it matches the given initial conditions.

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