Question

4. Find the general solution to 4y"+20y'+25y = 0

Answer 1

`y(x)=c_1e^{-(5)/(2)x}+c_2xe^{-(5)/(2)x}`

Explanation:

The given differential equation is 4y"+20y'+25y = 0

The characteristics equation is given by

`4r^2+20r+25=0`

Now, solve the equation for r

Factor by middle term splitting

`4r^2+10r+10r+25=0\\\\2r(2r+5)+5(2r+5)=0`

Factored out the common term

`(2r+5)(2r+5)=0`

Use Zero product property

`(2r+5)=0,(2r+5)=0`

Solve for r

`r_(1,2)=-(5)/(2)`

We got the repeated roots.

Hence, the general equation for the differential equation is

`y(x)=c_1e^{-(5)/(2)x}+c_2xe^{-(5)/(2)x}`