0
$\begingroup$

Suppose we wish to solve the second-order homogeneous differential equation ay″ + by′ + cy = 0, (3)

where a, b, and c are constants. To solve Equation (3), we seek a function which when multiplied by a constant and added to a constant times its first derivative plus a constant times its second derivative sums identically to zero.

One function that behaves this way is the
exponential function 

    y = e^rx

, when r is a constant.

This is how my textbook proceeds to solve the equation (3) and it works, but is y = e^rx the only function that can solve (3) ? Then why does everybody use y=e^rx ?

$\endgroup$
1
$\begingroup$

For $a=1,b=0,c=1$, the solution of the ODE is $C_1\cos x+ C_2 \sin x$. So $e^{rx}$ is not the only kind of solution.

$\endgroup$
  • $\begingroup$ no, we want a function like y = e^rx, that can be plugged into (3), which reduces (3) into its auxillary form $\endgroup$ – theenigma017 Apr 28 at 7:41
  • 1
    $\begingroup$ No it is not, that is used to check the roots of the characteristic equation. If they are distinct and real then it would be of the form $e^{r_1x}$ and $e^{r_2x}$, otherwise you have to choose different basis functions(Like I showed in my answer). One can show that if the wronskian of the basis function is nonzero, then the solution is unique. en.wikipedia.org/wiki/Wronskian. For the case of distinct and real roots of characteristic equation with constant coefficients, the solution is always unique. $\endgroup$ – user88923 Apr 28 at 8:04
  • $\begingroup$ But my textbook seems too use the same basis function for non-real roots and non-distinct roots $\endgroup$ – theenigma017 Apr 28 at 9:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.