Question

Find the general solution of the following differential equations:
(a) y''+6y'+5y=0
(b) y''+2y'+y=0
(c) y''+4y=0
(d) y''+4y'=0

Answer 1

I just studied this topic yesterday xD

(a) r^2 + 6r + 5 = 0
(r+5)(r+1) = 0
r = - 5 , r =- 1

\[y(x) = c_1e^{-5x} + c_2e^{-x}\]

Answer 2

You don't need to provide steps, I created those questions

Answer 3

because I learnt differential equations before but I forgot them

Answer 4

(b) r^2 + 2r + 1 = 0
(r+1)^2
r = -1
\[y(x) = c_1e^{-x} + c_2xe^{-x}\]

Answer 5

I just forgot how to change from aux equations to solution

Answer 6

oh, okay!
the formula for the general solution is:
\[ y = c_1e^{r_1x} + c_2e^{r_2x}\]
if you have two roots
if you one root it is:
\[y = c_1e^{r_1x} + c_2xe^{r_1x}\]
if you have a complex root you use euler's equation:
\[y = e^{\alpha x} ( c_1\cos \beta x + c_2 \sin \beta x)\]

where \[\alpha = -b/2a , \beta = \frac{ \sqrt{4ac-b^2} }{ 2a }\]

Answer 7

\[y'' = r^2 , y' = r , y = 1\]

Answer 8

Please just tell me the answer, I can learn from the answer, if I still don't understand I'll find you thank you :)

Answer 9

Okay.
(c) y'' + 4y = 0

r^2 + 4 = 0
\[r = \pm 2i\]

\[\alpha = 0 , \beta = 2\]

\[y = c_1 \cos 2x + c_2\sin 2x\]

Answer 10

(d) y'' + 4y' = 0
r^2 + 4r = 0
r(r+4) = 0
r = -4 ,r = 0
\[y = c_1 e^{-4x} + c_2x\]

Answer 11

@shamil98 Thank you so much <3

Answer 12

np :)

Answer 13

I wanna test what would happen if I regard +-2i as normal roots...

Answer 14

\[y(x)=\Large c_1e^{2ix}+c_2e^{-2ix}=c_1(\cos2x+i\sin2x)+c_2(\cos2x-i\sin2x)\\\Large(c_1+c_2)\cos2x+(c_1i-c_2i)\sin2x=c_3\cos2x+c_4\sin2x\]
Oh that makes sense now thank you :)