Find the general solution of DE 4y "+ y '= 0

y=a+be^(-1/4t)

Step-by-step explanation:

We can solve the characteristic equation, 4r^2+r=0 to help.

Factor:

r(4r+1)=0

This implies r=0 or 4r+1=0.

The second equation can be solved by subtracting 1 on both sides and then dividing boths sides by 4 like so:

4r+1=0

4r+1-1=0-1

4r=-1

4r/4=-1/4

r=-1/4

So r=0 and r=-1/4 which means the general solution is y=ae^(0t)+be^(-1/4t)

Simplifying gives y=a+be^(-1/4t) since e^0=1.

Let's check our solution.

y=a+be^(-1/4t)

y'=0+-1/4be^(-1/4t)

y''=0+1/16be^(-1/4t)

Plug in:

4y''+y'=0

4*1/16be^(-1/4t)+-1/4be^(-1/4t)=0

1/4be^(-1/4t)+-1/4be^(-1/4t)=0

0=0 is a true equation so the solution has been verified.

-10i^2-4i

step-by-step explanation:

evaluate for p=2,q=5i

(2−4)(2+5i)i

(2−4)(2+5i)i

=−10i2−4i