Answer:
Step-by-step explanation:
This is a second degree, homogeneous differential Equation.
Here’s how this works:
Set y = e^(rx) .
Then, y’ = r*e^(rx) and y’’ = r^2*(e^(rx)).
When I plug everything in, I get the following characteristic equation:
r^2 +3*r +7 = 0
Solving for r, I get
r1 = (-3 +i*sqrt(19))/2 and r2 = (-3-i*sqrt(19))/2
So, your solution will look like the following:
y = e^(-3/2*x) [A*cos(x*sqrt(19)/2) + B*sin(x*sqrt(19)/2)]
Now, you want to use the initial conditions to solve for A and B.
y(0) = 1 means when x =0, y = 1.
We know that A = 1.
y’(0) = 2 means when x = 0, y’ = 2.
To solve for B, you can take the derivative. It involves the product Rule. Replace x with 0 and y’ with 2. You’ll be able to find B. I hope this helps.
