I am studying about second order homogeneous equations with constant coefficients of the form:
L(y) = y'' + a1y' + a2y = 0
A possible solution to this would be e^rx, where r is a constant. By subbing this in, we find that this would be the solution if the characteristic polynomial p(r) (which is a quadratic in terms of r) equals zero.
In the case that the roots are equal, then p(r)=p'(r)=0.
In order to use this extra information that p'(r)=0 (to get another distinct solution), we differentiate our homogeneous equation w.r.t r, currently in the form: L(e^rx)=p(r)e^rx .. Here I have a problem with the differentiation on the left hand side. According to my text, the differentiation goes:
(as shown in image)
So apparently, we just ignore the function L, and directly differentiate e^rx with respect to r, inside the function L. The text says we do this, because L involves differentiation only with respect to x. I'm not getting how they differentiated here, I'm not able to apply any of the standard rules of differentiation (which i know of). Can someone explain this differentiation process a little better?
Best Answer
$$\partial_r (L(e^{rx}))=\partial_r (\partial_{xx}(e^{rx})+a_1\partial_x (e^{rx})+a_2e^{rx})$$ $$\partial_r (L(e^{rx}))=\partial_r \partial_{xx}(e^{rx})+a_1\partial_r \partial_x (e^{rx})+a_2\partial_r e^{rx}$$ You switch the order of derivatives: $$\partial_r (L(e^{rx}))=\partial_{xx}\partial_r(e^{rx})+a_1\partial_x \partial_r(e^{rx})+a_2\partial_re^{rx})$$ $$\partial_r (L(e^{rx})=L(\partial_r(e^{rx}))$$ Finally: $$\partial_r (L(e^{rx})=L(xe^{rx})$$