Substituting y1(t)=c0ekat into the differential equation for y2(t) (equation (2): dtdy2=kay1kcy2 we obtain dtdy2=kac0ekatkcy2 A test solution to this differential equation takes the form y2(t)=Aekat+Bekct where A and B are constants to be determined. Exercise: Differentiate the test solution above (4) to obtain dtdy2 and call this (5). Show your working. Answer: Exercise: Substitute the test solution (4) into the right hand side of the differential equation above (3). Show your working. Answer: Exercise: Set the derivative of the test solution (5) equal to the answer from the previous exercise so that you have an expression involving A,B,ekat,ekct on both left and right