Question

This question concerns the differential equation x(d^2y)/(dx^2)-(2x+1)(dy)/(dx)+(x+1)y=4xe^x, and the associated homogeneous differential equation x(d^2y)/(dx^2)-(2x+1)(dy)/(dx)+(x+1)y=0.

1. Show that y_1(x)=e^x is a solution of the homogeneous differential equation.
2. Use the method of reduction of order to show that a second linearly independent solution of the homogeneous differential equation is y_2(x)=x^2e^x.
3. Use the method of variation of parameters to find the general solution of the given nonhomogeneous differential equation [Hint: Write the differential equation in standard form!].

x(d^2y)/(dx^2)-(2x+1)(dy)/(dx)+(x+1)y=4xe^x.

1. Write down homogeneous equation:

x(d^2y)/(dx^2)-(2x+1)(dy)/(dx)+(x+1)y=0.

y_1(x)=e^x is solution of the given homogeneous equation because

x(d^2(y_1(x)))/(dx^2)-(2x+1)(d(y_1(x)))/(dx)+(x+1)y_1(x)=

=x(d^2(e^x))/(dx^2)-(2x+1)(d(e^x))/(dx)+(x+1)e^x=

=xe^x-(2x+1)e^x+(x+1)e^x=e^x(x-2x-1+x+1)=e^x*0=0.

2. Now, assume that second solution is y_2(x)=v(x)y_1(x)=v(x)e^x.

Then (dy_2)/(dx)=(dv)/(dx)e^x+ve^x and

(d^2y_2)/(dx^2)=(d^2v)/(dx^2)e^x+(dv)/(dx)e^x+(dv)/(dx)e^x+ve^x=(d^2v)/(dx^2)e^x+2(dv)/(dx)e^x+ve^x.

Plugging these results into the given homogeneous differential equation gives:

x((d^2v)/(dx^2)e^x+2(dv)/(dx)e^x+ve^x)-(2x+1)((dv)/(dx)e^x+ve^x)+(x+1)ve^x=0

Or

x((d^2v)/(dx^2)e^x-(dv)/(dx)e^x)=0.

Divide left and right parts by xe^x:

(d^2v)/(dx^2)-1/x(dv)/(dx)=0.

Now, let w=(dv)/(dx) then (dw)/(dx)=(d^2v)/(dx^2) and equation can be rewritten as

(dw)/(dx)-1/xw=0.

(dw)/w=(dx)/x.

int(dw)/w=int(dx)/x.

ln(w)=ln(x)+c, where c is arbitrary constant.

Now,

e^(ln(w))=e^(ln(x)+c).

w=e^cx=c_1x, where c_1 is arbitrary constant.

Since w=(dv)/(dx) then (dv)/(dx)=c_1x.

v=intc_1xdx=c_1/2x^2+c_2.

Since we need any particular solution for v then take c_1=2, c_2=0.

Thus, v=x^2.

Finally, y_2(x)=ve^x=x^2e^x.

Clearly y_1(x) and y_2(x) are linearly independent.

So, general solution of the given homogeneous differential equation is

y_h=a_1e^x+a_2x^2e^x, where a_1 and a_2 are arbitrary constants.

3. Rewrite the given inhomogeneous equation in standard form. For this divide left and right part by x:

(d^2y)/(dx^2)-((2x+1))/x(dy)/(dx)+((x+1))/xy=4e^x.

Now assume that constants in the homogeneous solution are functions of x. Set up following system:

{((da_1)/(dx)e^x+(da_2)/(dx)x^2e^x=0), ((da_1)/(dx)(d(e^x))/(dx)+(da_2)/(dx)(d(x^2e^x))/(dx)=4e^x):}.

Or

{((da_1)/(dx)e^x+(da_2)/(dx)x^2e^x=0),((da_1)/(dx)e^x+(da_2)/(dx)(2xe^x+x^2e^x)=4e^x):}.

Subtract first equation from second:

(da_1)/(dx)e^x+(da_2)/(dx)(2xe^x+x^2e^x)-((da_1)/(dx)e^x+(da_2)/(dx)x^2e^x)=4e^x-0

Which yields

2xe^x(da_2)/(dx)=4e^x

Or

(da_2)/(dx)=2/x.

a_2=int2/xdx=2ln(x).

(We've taken integration constant 0, since we need particular solution).

Now, from first equation of the system

(da_1)/(dx)=-(da_2)/(dx)x^2=-2/x*x^2=-2x.

a_1=int-2xdx=-x^2.

(We've taken integration constant 0, since we need particular solution).

Therefore,

y_p=a_1(x)e^x+a_2(x)x^2e^x=-x^2e^x+2ln(x)x^2e^x.

Finally general solution of the given inhomogeneous system is

y=y_h+y_p=a_1e^x+a_2x^2e^x-x^2e^x+2ln(x)x^2e^x

Or

y=a_1e^x+a_2x^2e^x+x^2e^x(2ln(x)-1).