## anonymous 5 years ago 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!

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