Question

Question #76309

Answer 1

For homogenous linear differential equation with constant coefficients, the general solution has the form

`y(t) = C e^(lambda t)`. Substituing this solution into the homogeneous equation, we have

`(a_n lambda^n+a_(n-1)lambda^(n-1)+cdots+a_1lambda + a_0)Ce^(lambda t) = 0`

`e^(lambda t ) ne 0` so the equation is verified for `lambda's` which are the roots of

`p(lambda) = a_n lambda^n+a_(n-1)lambda^(n-1)+cdots+a_1lambda + a_0`

the so called characteristic polynomial.

If `lambda_k, k=1,2,cdots,lambda_n` are the roots, supposing that they are distinct, them the general solution will be posed as

`y(t) = sum_k C_k e^(lambda_k t)`

In case of repeated roots, appear associated polinomials in `t` with their degree equal to the repetition level.

The constants `C_k` are determined according with the initial conditions.

https://en.wikipedia.org/wiki/Linear_differential_equation

Answer 2

See explanation.

Explanation:

It is found that the differential equation can be converted to a

polynomial equation, with the same coefficients, by the substitution

`y=e^(rx)`, using `n^(th)` derivative of `e^(rx)=r^ky, k =1, 2, 3, ...,n`.

With D for the differentiation operator `d/(dx)`, the conversion

gives.

`sum a_kD^ky=(sum a_kr^k)e^(rx)=0`

As `e^(rx)>0,sum a_kr^k=0`

It follows that , for every root `r_k` of this polynomial equation,

`y = e^(r_kx)` is a solution for the the differential equation.

Also, an arbitrary scalar `A_k xx e^(r_kx)` is a solution..

Theoretically, reduction of the order of the differential equation by

every integration produces one constant of integration. So,

successive integration n times to produce the general solution

would deposit n constants of integration.

Now, the linear sum

`y=sum A_ke^(r_kx)` substituted in the differential equation leads to

`sum (A_k)(0)=0`.

And so, we are justified in stating that

`y=sum A_ke^(r_kx)` is the general solution.

There are particular cases like `D^3y=0`, wherein this method

leads to the degenerate case.

Here, `y=ax^2+bx +c` is a quadratic in x

Here, 0 is a thrice repeated root. of the characteristic equation.

For `(D^3+D^2)y=D^2(D+1)y=0`, it is a combination

`y=ax+b+ce^(-x)`, against the three roots 0, 0 and -1.

The part ax + b comes from direct integration, twice in succession,

for removing D^2. The other operator D+1 gives the part `ce^(-x)`

Substitute separately both in the differential equation and see

how it works.

.