`lambda^2+4lambda+4=0`

Solution is `(lambda+2)^2 =0`

`lambda = -2`

We then obtain `u_1=e^(-2x)` and `u_2 = xe^(-2x)`

The Wronskian is

`W=-e^(-2x)e^(-2x)(2x-1)+2xe^(-4x)=e^(-4x)`

Since the Wronksian is non zero the two functions are independent and therefore the two functions are the general solution to the differential equation.

We find the particular solution by...

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`lambda^2+4lambda+4=0`

Solution is `(lambda+2)^2 =0`

`lambda = -2`

We then obtain `u_1=e^(-2x)` and `u_2 = xe^(-2x)`

The Wronskian is

`W=-e^(-2x)e^(-2x)(2x-1)+2xe^(-4x)=e^(-4x)`

Since the Wronksian is non zero the two functions are independent and therefore the two functions are the general solution to the differential equation.

We find the particular solution by computing `A(x)u_1+B(x)u_2`

`A(x)=-int 1/Wu_2(x)b(x) dx = -int 1/e^(-4x) xe^(-2x) e^(-2x)/x^2 dx=-int 1/x dx`

So `A(x) = -ln(x)+C_1`

`B(x) = int 1/W u_1(x)b(x) dx = int 1/e^(-4x)e^(-2x)e^(-2x)/x^2 dx = int 1/x^2 dx`

So `B(x)=-1/x + C_2`

So our general solution is

`y = (-ln(x)+C_1)e^(-2x)+(-1/x+C_2)xe^(-2x)` Simplifying

`y = -(ln(x)+1)e^(-2x) + C_1e^(-2x) + C_2xe^(-2x)`