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

Answer:

Therefore the particular solution of the given differential equation is  

$Y(x)=\frac{1}{6} x^2 ln x$

Step-by-step explanation:

The given ordinary differential equation is

$x^2y"-3xy' +4y=11x^2$

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.

$y'_2=2x ln x+ x^2\times\frac{1}{x}$

    $= 2x ln x+x$

$y"_2(x)=2x\times \frac{1}{x} +2lnx+1$ = 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 $g(x)=\frac{y_2(x)}{y_2(x)} = ln x$

The particular solution is

$Y(x)=- y_1(x)\int\frac{y_2(x).g(x)}{W(y_1,y_2)(x)}dx + y_2(x)\int\frac{y_1(x).g(x)}{W(y_1,y_2)(x)}dx$

      $=-x^2\int\frac{x^2 ln x . ln x}{x^3} dx+x^2ln x\int \frac{x^2ln x}{x^3} dx$                

      $=-x^2\int \frac{(lnx)^2}{x} dx+x^2 ln x\int \frac{ln x}{x} dx$

       $=-x^2\int u^2 du+x^2ln x\int u dx$                       Let ln x  =u   $\Rightarrow \frac{1}{x} dx=du$

      $=-x^2 \frac{u^3}{3} +x^2 ln x \frac{u^2}{2}$

        $=-\frac{x^2(lnx)^3}{3} +\frac{x^2(lnx)^3}{2}$

       $=\frac{1}{6}x^2 (ln x)^3$

Therefore the particular solution of the given differential equation is  

$Y(x)=\frac{1}{6} x^2 ln x$