Solve appropriate linear differential equations using techni
Solution
We can rewrite the ode as
x^2y\'\'-2y=x
First we solve the associated homogeneous ode ie
x^2y\'\'-2y=0
LEt, y=x^m
Substituting gives
m(m-1)x^m-2x^m=0
m(m-1)-2=0
m^2-m-2=0
m^2-2m+m-2=0
m(m-2)+(m-2)=0
m=-1,2
HEnce, 1/x and x^2 are solutions to this homogeneous ode
So general solution to homogeneous ode is
yh(x)=A/x+Bx^2
Now we look for particular solution
Base on inhomogeneous part ie x
We make a guess for particular solution
yp=Ax
Substituting gives
-2Ax=x
A=-1/2
yp=-x/2
So general solution is
y(x)=yh+yp=A/x+Bx^2-x/2
