Question #76309

2 Answers
Sep 20, 2016

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

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.

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