Example 2Video ExampleFind the solution of the initial value problem.x2y'+xy=4,x>0,y(1)=5SolutionWe must first divide both sides by the coefficient ofy'to put the differential equation into standard form.y'+1xy=4x2,x>0The integration factor isI(x)=e(1x)dx=eln(x)=Multiplication of(1)byx givesxy'+y=4xor(,)'=4xThenxy=4xdx=,+Cand soy=+CxSince y(1)=5,we have 5=4ln(1)+C1=Therefore, the solution to the initial value problem isy=