Question

d^2/dx^2 - dy / dx -12y =0 y(0)=3 y' (0) =5​

Answer 1

Answer:

The answer of the question is y(x) = 2e^4x+1e^(-3)x

Step-by-step explanation:

The ordinary differential equation is,

(D^2-D-12)y=0

Let y=e^mx be a solution of the equation, then

m^2-m-12=0

So, m=4, -3

So the complementary function is, y(x) =c1e^4x+c2e^(-3)x

And y'(x) =4c1e^4x-3c2e^(-3)x

Now putting the value of y(0) and y'(0) we will get two equation,

c1+c2=3 and 4c1-3c2=5

Solving them we will get, c1=2 and c2 =1

So, y(x) =2e^4x+1e^-3x