Second order nonhomogeneous using wronskian: x^2 y" - 3xy' + 4y = log(x)

[devinaxx]

There is a second-order non-homogeneous differential equation, as follows:

. . . . .$\displaystyle x^2 y"\, -\, 3xy'\, +\, 4y\, =\, \log(x)$

I have to find the solution. Here is my attempt:

First, I found the solution to the homogeneous equation. It is a repeated root, so:

. . . . .$\displaystyle yp\, =\, Ax\ln(x)\, +\, Bx^2$

Here for the non-homogeneous equation, I used variation parameter:

. . . . .$\displaystyle y_1\, =\, x^2 \ln(x)$

. . . . .$\displaystyle y_2\, =\, x^2$

Using the Wronskian and Cramer's Rule:

. . . . .$\displaystyle yp\, =\, u_1.y_1\, +\, u_2.y_2$

. . . . .$\displaystyle W\, =\, \left[\begin{matrix}x^2 \ln(x) & x^2 \\ 2x \ln(x)\, +\, x & 2x\end{matrix}\right]\, =\, -x^3$

. . . . .$\displaystyle W_1\, =\, \left[\begin{matrix}0 & x^2 \\ \log(x) & 2x\end{matrix}\right]\, =\, -x^2 \ln(x)$

. . . . .$\displaystyle W_2\, =\, \left[\begin{matrix}x^2 \ln(x) & 0 \\ 2x \ln(x)\, +\, x & \log(x)\end{matrix}\right]\, =\, x^2 \ln^2(x)$

. . . . .$\displaystyle u'\, =\, \dfrac{W_1}{W}\, \dfrac{-x^2 \ln(x)}{-x^3}\, =\, \dfrac{\ln(x)}{x}$

. . . . .$\displaystyle u'\, =\, \dfrac{W_2}{W}\, \dfrac{-x^2 \ln^2(x)}{-x^3}\, =\, \dfrac{-\ln^2(x)}{x}$

I integrated to find u₁ and u₂:

. . . . .$\displaystyle u_1\, =\, \frac{1}{2} \ln^2(x)$

. . . . .$\displaystyle u_2\, =\, -\frac{1}{3} \ln^3(x)$

Then:

. . . . .$\displaystyle yp\, =\, u_1.y_1\, +\, u_2.y_2$

. . . . . . . .$\displaystyle =\, \frac{1}{2} \ln^2(x).x^2 \ln(x)\, -\, \frac{1}{3} \ln^3(x).x^2\, =\, \frac{1}{6} x^2 \ln^3(x)$

Is this correct?

By the way, the book says that the answer is:

. . . . .$\displaystyle y\, =\, x^2 \left(A \log(x)\, +\, B\right)\, +\, \frac{1}{?} \left(\log(x)\, +\, 1\right)$

Why this is not

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[mmm4444bot]

There is a second-order non-homogeneous differential equation, as follows:

. . . . .$\displaystyle x^2 y"\, -\, 3xy'\, +\, 4y\, =\, \log(x)$

I have to find the solution. Here is my attempt:

First, I found the solution to the homogeneous equation. It is a repeated root, so:

. . . . .$\displaystyle yp\, =\, Ax\ln(x)\, +\, Bx^2$

Here for the non-homogeneous equation, I used variation parameter:

. . . . .$\displaystyle y_1\, =\, x^2 \ln(x)$

. . . . .$\displaystyle y_2\, =\, x^2$

Using the Wronskian and Cramer's Rule:

. . . . .$\displaystyle yp\, =\, u_1.y_1\, +\, u_2.y_2$

. . . . .$\displaystyle W\, =\, \left[\begin{matrix}x^2 \ln(x) & x^2 \\ 2x \ln(x)\, +\, x & 2x\end{matrix}\right]\, =\, -x^3$

. . . . .$\displaystyle W_1\, =\, \left[\begin{matrix}0 & x^2 \\ \log(x) & 2x\end{matrix}\right]\, =\, -x^2 \ln(x)$

. . . . .$\displaystyle W_2\, =\, \left[\begin{matrix}x^2 \ln(x) & 0 \\ 2x \ln(x)\, +\, x & \log(x)\end{matrix}\right]\, =\, x^2 \ln^2(x)$

. . . . .$\displaystyle u'\, =\, \dfrac{W_1}{W}\, \dfrac{-x^2 \ln(x)}{-x^3}\, =\, \dfrac{\ln(x)}{x}$

. . . . .$\displaystyle u'\, =\, \dfrac{W_2}{W}\, \dfrac{-x^2 \ln^2(x)}{-x^3}\, =\, \dfrac{-\ln^2(x)}{x}$

I integrated to find [FONT=&quot]u₁[/FONT] and [FONT=&quot]u₂[/FONT]:

. . . . .$\displaystyle u_1\, =\, \frac{1}{2} \ln^2(x)$

. . . . .$\displaystyle u_2\, =\, -\frac{1}{3} \ln^3(x)$

Then:

. . . . .$\displaystyle yp\, =\, u_1.y_1\, +\, u_2.y_2$

. . . . . . . .$\displaystyle =\, \frac{1}{2} \ln^2(x).x^2 \ln(x)\, -\, \frac{1}{3} \ln^3(x).x^2\, =\, \frac{1}{6} x^2 \ln^3(x)$

Is this correct?

By the way, the book says that the answer is:

. . . . .$\displaystyle y\, =\, x^2 \left(A \log(x)\, +\, B\right)\, +\, \frac{1}{?} \left(\log(x)\, +\, 1\right)$

Why this is not

Oops -- something happened to the rest of your question

Note: I'm using symbol ln(x), for the Natural log of x.

Your solution is

y(x) = 1/6 * x^2 * ln(x)^3

Therefore

y'(x) = 1/3 * x * ln(x)^3 + 1/2 * x * ln(x)^2

and

y''(x) = 1/3 * ln(x)^3 + 3/2 * ln(x)^2 + ln(x)

If we substitute these expressions into the left-hand side of the equation

x^2 * y''(x) - 3 * x * y'(x) + 4 * y(x) = ln(x)

it does not simplify to ln(x). It simplifies to x^2 * ln(x), instead.

I would start by double-checking your steps (I did not). I can confirm that the correct solution fits the book's form

y(x) = x^2 * [C₁ * ln(x) + C₂] + 1/N * [ln(x) + 1]