Answer to Question #58420 in Differential Equations for Joseph

Question

6 The solution to the differential equation
(4+t2)dydt+2ty=4t
is
y=4t24+t2+c6+t2
y=2t34+t3+c4+t2
y=2t27+t2
y=2t24+t2+c4+t2
7 Find the general solution of the differential equation
dydx−2y=4−x
y=−12+23x+ce2x
y=−74+12x+ce2x
y=−74+22x+ce3x
y=−ln74+12x+celnx
8 Solve the initial value problem
xdydx+2y=4x2
,
y(1)=2
y=x2+1x2,x>0
y=3x2+1x2,x>0
y=x−2+1x2,x>0
y=x3+13x2,x>0

9 Solve for
dydx+y=5sin2x
y=ce2x−sin2x−2cos3x
y=ce−x+tan2x−2cos2x
y=ce−x+sin2x+3cos2x
y=ce−x+sin2x−2cos2x

10 Given the differential equation
dydx+(1x)y=3cos2x,x>0
the solution is __
y=cx+3cos2x4x+3sin22
y=cx+3cos2x4x
y=cx+3sin22
y=c2x+3cosx4x+3sin62