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General solution to a homogeneous equation

If y 1 ( x ) and y 2 ( x ) are linearly independent solutions to a second-order, linear, homogeneous differential equation, then the general solution is given by

y ( x ) = c 1 y 1 ( x ) + c 2 y 2 ( x ) ,

where c 1 and c 2 are constants.

When we say a family of functions is the general solution to a differential equation , we mean that (1) every expression of that form is a solution and (2) every solution to the differential equation can be written in that form, which makes this theorem extremely powerful. If we can find two linearly independent solutions to a differential equation, we have, effectively, found all solutions to the differential equation—quite a remarkable statement. The proof of this theorem is beyond the scope of this text.

Writing the general solution

If y 1 ( t ) = e 3 t and y 2 ( t ) = e −3 t are solutions to y 9 y = 0 , what is the general solution?

Note that y 1 and y 2 are not constant multiples of one another, so they are linearly independent. Then, the general solution to the differential equation is y ( t ) = c 1 e 3 t + c 2 e −3 t .

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If y 1 ( x ) = e 3 x and y 2 ( x ) = x e 3 x are solutions to y 6 y + 9 y = 0 , what is the general solution?

y ( x ) = c 1 e 3 x + c 2 x e 3 x

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Second-order equations with constant coefficients

Now that we have a better feel for linear differential equations, we are going to concentrate on solving second-order equations of the form

a y + b y + c y = 0 ,

where a , b , and c are constants.

Since all the coefficients are constants, the solutions are probably going to be functions with derivatives that are constant multiples of themselves. We need all the terms to cancel out, and if taking a derivative introduces a term that is not a constant multiple of the original function, it is difficult to see how that term cancels out. Exponential functions have derivatives that are constant multiples of the original function, so let’s see what happens when we try a solution of the form y ( x ) = e λ x , where λ (the lowercase Greek letter lambda) is some constant.

If y ( x ) = e λ x , then y ( x ) = λ e λ x and y = λ 2 e λ x . Substituting these expressions into [link] , we get

a y + b y + c y = a ( λ 2 e λ x ) + b ( λ e λ x ) + c e λ x = e λ x ( a λ 2 + b λ + c ) .

Since e λ x is never zero, this expression can be equal to zero for all x only if

a λ 2 + b λ + c = 0 .

We call this the characteristic equation of the differential equation.


The characteristic equation    of the differential equation a y + b y + c y = 0 is a λ 2 + b λ + c = 0 .

The characteristic equation is very important in finding solutions to differential equations of this form. We can solve the characteristic equation either by factoring or by using the quadratic formula

λ = b ± b 2 4 a c 2 a .

This gives three cases. The characteristic equation has (1) distinct real roots; (2) a single, repeated real root; or (3) complex conjugate roots. We consider each of these cases separately.

Distinct real roots

If the characteristic equation has distinct real roots λ 1 and λ 2 , then e λ 1 x and e λ 2 x are linearly independent solutions to [link] , and the general solution is given by

y ( x ) = c 1 e λ 1 x + c 2 e λ 2 x ,

where c 1 and c 2 are constants.

For example, the differential equation y + 9 y + 14 y = 0 has the associated characteristic equation λ 2 + 9 λ + 14 = 0 . This factors into ( λ + 2 ) ( λ + 7 ) = 0 , which has roots λ 1 = −2 and λ 2 = −7 . Therefore, the general solution to this differential equation is

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Practice Key Terms 7

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Source:  OpenStax, Calculus volume 3. OpenStax CNX. Feb 05, 2016 Download for free at http://legacy.cnx.org/content/col11966/1.2
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