Given a second order linear homogeneous differential equation a2(x)y′′+a1(x)y′+a0(x)y=0 we know that a fundamental set for this ODE consists of a pair linearly independent solutions y1,y2. But there are times when only one function, call it y1, is available and we would like to find a second linearly independent solution. We can find y2 using the method of reduction of order. First, under the necessary assumption the a2(x)≠0 we rewrite the equation as y′′+p(x)y′+q(x)y=0 p(x)=a1(x)a2(x), q(x)=a0(x)a2(x), Then the method of reduction of order gives a second linearly independent solution as y2(x)=Cy1u=Cy1(x)∫e−∫p(x)dxy21(x)dx where C is an arbitrary constant. We can choose the arbitrary constant to be anything we like. One useful choice is to choose C so that all the constants in front reduce to 1. For example, if we obtain y2=C3e2x then we can choose C=1/3 so that y2=e2x. Given the problem y′′−4y′+29y=0 and a solution y1=e2xsin(5x) Applying the reduction of order method we obtain the following y21(x)=Cy1u= e^(4x)sin^2(5x) p(x)= -4 and e−∫p(x)dx= x So we have ∫e−∫p(x)dxy21(x)dx=∫ 1 dx= x Finally, after making a selection of a value for C as described above (you have to choose some nonzero numerical value) we arrive at y2(x)=Cy1u= So the general solution to y′′−5y′+4y=0 can be written as y=c1y1+c2y2=c1 +c2