4y''-y=0 Differential Equation y(-2)=2 and y'(-2)=-1?

Characteristic eq is 4r^2-1=0
r=1/2 and -1/2

y=c1e^(t/2)+c2e^(-t/2)

after plugging in the initial values, i still could not solve for c1 and c2!
Please help me find c1 and c2!

Question

4y''-y=0 Differential Equation y(-2)=2 and y'(-2)=-1?

Characteristic eq is 4r^2-1=0
r=1/2 and -1/2

y=c1e^(t/2)+c2e^(-t/2)

after plugging in the initial values, i still could not solve for c1 and c2!
Please help me find c1 and c2!

anonymous · Answer

Applying the given conditions:
$$2=c_1e^{-1}+c_2e... (1)$$
$$-1={1 \over 2}c_1e^{-1}-{1 \over 2}c_2e \implies -2=c_1e^{-1}-c_2e ... (2)$$
Add (1) and (2), you get that c1=0 and hence c2=2/e