Homogeneous linear equations 2

Calculus Level 1

What is the general solution to the differential equation y ′ ′ + 4 y ′ + 4 y = 0 y''+4y'+4y=0 ?

c 1 e − 2 x c_1e^{-2x} c 1 e − 2 x + c 2 x e − 2 x c_1e^{-2x}+c_2xe^{-2x} ( c 1 + c 2 ) e − 2 x (c_1+c_2)e^{-2x} c 1 e 2 x + c 2 x e 2 x c_1e^{2x}+c_2xe^{2x}

Samir Khan by Samir Khan , 23, USA

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Samir Khan
Jun 16, 2016

The characteristic equation is r 2 + 4 r + 4 = 0 ⟹ r = − 2 r^2+4r+4=0\implies r=-2 , so the general solution is y = c 1 e − 2 x + c 2 x e − 2 x y=c_1e^{-2x}+c_2xe^{-2x} .

4 Helpful 2 Interesting 2 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...