Question

In Problems 21–24 verify that the indicated family of functions is a solution of the given differential equation. Assume an appropriate interval I of definition for each solution.

$x^3\frac{d^{3}y}{d{x}^3}+2x^2\frac{d^{2}y}{d{x}^2}-x\frac{dy}{dx}+y=12x^2;y=c_1{x}^{-1}+c_{2}x+c_{3}xlnx+4x^2$

Short answer

The indicated function is a solution of the given differential equation with the interval $I:(0,\infty )$.

Step-by-step solution

Determine the derivatives of the function.

Let the given function be $y=c_1{x}^{-1}+c_{2}x+c_{3}xlnx+4x^2$.

Then, the first derivative of the function is,

$\frac{dy}{dx}=-c_1{x}^{-2}+c_2+c_3(lnx+1)+8x$

The second derivative of the function is,

$\frac{d^{2}y}{d{x}^2}=2c_1{x}^{-3}+c_3{x}^{-1}+8$

The third derivative of the function is,

$\frac{d^{3}y}{d{x}^3}=-6c_1{x}^{-4}-c_3{x}^{-2}$

Determine the interval of the solution.

Substitute$y$,$y\text{'}$ and $y"$into the left-hand side of the differential equation.

$$\begin{gathered} x^3\left[-6c_1{x}^{-4}-c_3{x}^{-2}\right]+2x^2\left[2c_1{x}^{-3}+c_3{x}^{-1}+8\right]-x\left[-c_1{x}^{-2}+c_2+c_3(lnx+1)+8x\right]+\left[c_1{x}^{-1}+c_{2}x+c_{3}xlnx+4x^2\right]=12x^2 \\ {c}_1\left(-6x^{-1}+4x^{-1}+x^{-1}+x^{-1}\right)+c_2(-x+x)+c_3(-x+2x-xlnx-x+xlnx)+\left(16x^2-8x^2+4x^2\right)=12x^2 \\ 12x^2=12x^2 \end{gathered}$$

That is same as the right-hand side of the differential equation. As the solution has logarithmic term, the solution is defined when $x>0$, (i.e.) $I:(0,\infty )$. Thus, the indicated function is a solution of the given differential equation.