Question

Verify that the given functions y1 and y2 satisfy the corresponding homogeneous equation; then find a particular solution of the given nonhomogeneous equation. x2y'' − 3xy' + 4y = 11x2 ln x, x > 0; y1(x) = x2, y2(x) = x2 ln x

Answer 1

Therefore the particular solution of the given differential equation is

`Y(x)=(1)/(6) x^2 ln x`

Explanation:

The given ordinary differential equation is

`x^2y`

If y₁(x) =x² is a solution of ODE then it will be satisfy the ODE

y₁'(x)= 2x and y₁"(x)=2

Putting the value of y₁'(x) and y₁"(x) in x²y"-3xy'+4y=0 we get

x².2-3x.2x+4x²= 2x²-6x²+4x²=0

Therefore y₁(x) is a solution of the given differential equation.

Again,

y₂(x) =(x²ln x ) is a solution of ODE then it will be satisfy the ODE.

`y'_2=2x ln x+ x^2*(1)/(x)`

`= 2x ln x+x`

`y` = 3+2 ln x

Putting the value of y₂'(x) and y₂"(x) in x²y"-3xy'+4y=0 we get

`x^2 (3+2 ln x)-3x( 2x ln \ x+x)+4x^2\ ln\ x` =0

Therefore y₂(x) is a solution of the given differential equation.

The wronskian of y₁(x) and y₂(x) is

`W(y_1,y_2)(x)=\left|\begin{array}{cc}y_1&y_2\\y'_1&y'_2\end{array}\right|`

`=\left|\begin{array}{cc}x^2&x^2 ln x\\2x&2x ln x+x\end{array}\right|`

=x²(2x ln x+x)-x²ln x(2x)

=2x³ ln x +x³ - 2x³ln x

=x³≠0

Here
`g(x)=(y_2(x))/(y_2(x)) = ln x`

The particular solution is

`Y(x)=- y_1(x)\int(y_2(x).g(x))/(W(y_1,y_2)(x))dx + y_2(x)\int(y_1(x).g(x))/(W(y_1,y_2)(x))dx`

`=-x^2\int(x^2 ln x . ln x)/(x^3) dx+x^2ln x\int (x^2ln x)/(x^3) dx`

`=-x^2\int ((lnx)^2)/(x) dx+x^2 ln x\int (ln x)/(x) dx`

`=-x^2\int u^2 du+x^2ln x\int u dx` Let ln x =u
`\Rightarrow (1)/(x) dx=du`

`=-x^2 (u^3)/(3) +x^2 ln x (u^2)/(2)`

`=-(x^2(lnx)^3)/(3) +(x^2(lnx)^3)/(2)`

`=(1)/(6)x^2 (ln x)^3`

Therefore the particular solution of the given differential equation is

`Y(x)=(1)/(6) x^2 ln x`