Finding Linear Homogeneous Differential Equations With the General Solution

definition

The definition is easy to understand, just look at the highest order derivative. The difference between homogeneous and non-homogeneous depends on whether the term without y on the right is zero
Next, let's change the way of expression, generally better to study:

Second-order homogeneous linear differential equation

Solution superposition

Such processing can make the form of the solution more concise, and it is also easier to understand the principle of superposition

Linear correlation is not linear

Note: The knowledge of linear algebra is used below, but I learned this by myself last semester, and there may be errors.

Just remember two criteria to judge linear correlation:
1. One of a group of things can be expressed by other things, then this group is linearly related. On the contrary, if a set of coefficients that are not all zero can not be found so that one can be expressed by the other (it must be constant), It is linearly independent. From a geometric point of view, whether a set of vectors can be used as a basis, if there is excess, it is linearly related, and if there is no excess, it is linearly independent.
2. Groups with zeros are linearly related

General problem

That is to say, for an n-th order differential equation, we select n linearly independent special solutions, and use them to express all the solutions linearly. Then the reason, the teacher didn't talk about it in class, and I couldn't figure out how to prove it. Then I built a model to help understand. This model is not rigorous, because there is such a solution space, its dimension and the order of the differential equation. Same, and then each solution is a vector, so can we find n linearly independent special solutions to represent all solutions

Second-order non-homogeneous linear differential equation

This thing has appeared in linear algebra, so I will make some unwarranted conjectures. A differential equation is a system of linear equations with the same number of variables as its order. Then the general solution of the equation is the general solution of its derived group plus a special feature. solution.

Solution superposition theorem

This is obvious with matrix one, and then expand it

This can be explained from the structure of the solution after calculation (the tuft in the back is still y0, don't care about it in the front) or the principle of superposition of the solution (f(x) in the back)

In this case, the two solutions are subtracted by one, and y0 is eliminated at the end, and the front is not zero (because these are two different solutions, the Y part in front cannot be the same, so after the subtraction, at least one special solution of the derived group is left )

This seckill depends on the coefficient of y0, and all the ones that are not 1 are excluded. Okay, choose D
To prove why D is a general solution, the teacher's approach is to assume that y1-y3 and y2-y3 are linearly related, and then there is this formula. After simplification, it is easy to get that the three preceding coefficients cannot be all zero at all. And y1y2y3 can be moved (so we fix y1y2, and then let y3 fly, which is definitely not control), so these two are linearly independent, and then these two are the special solutions of the derived group, so Passed

But it feels not intuitive enough, and then I didn't come up with an intuitive solution

For this question, find a general solution of a second-order non-homogeneous differential equation. We have to find a special solution of the original equation and two special solutions of linearly independent derived groups. The previous one gives three and just choose one. The key is to derive two special solutions of the group. According to the structure of the general solution, find two and subtract one, and find that they are just linearly independent. That's right, it is these two. Finally, substitute the initial conditions to get the coefficients.

This problem is also very simple. Subtract one to obtain two linearly independent special solutions of the derived group, substitute in to solve the derived equation, and then pick a special solution to substitute in to obtain f(x). This is better than directly building three equations, because the solutions Homogeneous must be better than non-homogeneous

Second-order homogeneous linear differential equation with constant coefficients

The above are all explained relations, here you can really solve the equation

For this form, we guessed that the structure of e^(rx) can definitely be a special solution to be determined.
Then simplified to get a quadratic equation about r, the following three cases
Δ>0, direct two solutions, nothing more;
Δ=0, there is only one, you have to make up another one,

The common method is to replace the coefficient with the formula about x, substituting it and looking at it, and find that the last two items are missing, and the solution is directly obtained. Then here is to get the special solution, so take x directly.
Δ<0, there are only two complex roots, it is not good to write the complex root directly, please deal with it

Just use Euler's formula and then come up with two special solutions for real numbers. It's convenient to memorize the formula directly, and you can forget to push it now. If you push first, just use Euler's formula.

Generalized to high-order linear differential equations with constant coefficients

According to the basic theorem of algebra, an equation of degree n must have n roots in the complex number field, and then there are several roots (including multiple roots), which means that several linearly independent special solutions can be found (see the number of variable coefficients), and then The two changing laws are that the number of x increases.

In these two examples, the characteristic equations are first obtained, and then the solvable ones are solved. After dealing with multiple roots and complex roots, they can be combined linearly at the end.

Observe the roots so that you don't care about the coefficients and just look at the form. e^x has real roots, and e^(cos …sin) has complex roots. When you get what the root is, look at the number of x to get how many roots you have, and finally write it out Characteristic equation.

Second-order linear differential equation with constant coefficients

Is to get the general solution of the derived group first and then find a special solution
The key is how to find special solutions
Look at some special f(x)

In fact, if the title gives the actual number, just substitute it directly into that form to see the number of x on both sides and set it up. If there are parameters, pay attention to whether the two lumps are zero.

This is simple, don't talk about it

It took almost two hours and 40 minutes. I didn't expect this to be sorted out for so long. I will wait a little bit and come back tomorrow.