### Author Topic: Q4 TUT 0201  (Read 1484 times)

#### Victor Ivrii

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##### Q4 TUT 0201
« on: October 26, 2018, 05:41:18 PM »
Verify that the given functions $y_1$ and $y_2$ satisfy the corresponding homogeneous equation; then find a particular solution of the given inhomogeneous equation.
\begin{gather*}
x^2y'' - 3xy' + 4y = x^2 \ln (x),\qquad x > 0;\\
y_1(x) = x^2,\quad y_2(x) = x^2 \ln (x).
\end{gather*}

#### Pengyun Li

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##### Re: Q4 TUT 0201
« Reply #1 on: October 26, 2018, 05:41:52 PM »
Let $L(y) = x^2 y'' - 3xy' + 4y$

$y_1(x) = x^2, y_1'(x) = 2x, y_1''(x)= 2$

Sub it into $L(y)$, then $L(y) = 2x^2 - 3x(2x) + 4x^2 = 0$, hence $y_1(x) = x^2$ satisfies the homo equation.

$y_2(x) = x^2 ln(x), y_2'(x) = x + 2xln(x), y_2''(x) = 3 + 2ln(x)$

Sub it into $L(y)$, then $L(y) = x^2(3 + 2ln(x)) - 3x(x + 2xln(x) + 4x^2 ln(x) = 0$, hence $y_2(x) = x^2 ln(x)$ satisfies the homo equation.

$W[y_1, y_2](x) = \left|\begin{matrix}y_1(x) & y_2(x) \\ y_1'(x) & y_2'(x)\end{matrix}\right|= \left|\begin{matrix}x^2 & x^2 ln(x) \\ 2x & x + 2xln(x)\end{matrix}\right| = x^3$

Let the particular solution $y_p(x) = u_1 y_1(x) + u_2 y_2(x)$,

Since $x^2 y'' - 3xy' + 4y = x^2 ln(x)$, thus $y'' - \frac{3}{x}y' + \frac{4}{x^2}y = ln(x)$

$p(x) = - \frac{3}{x}, q(x) = \frac{4}{x^2}, g(x) = ln(x)$

$u_1 = -\int{\frac{y_2(x)g(x)}{W[y_1, y_2](x)}}dx = - \int{\frac{x^2 ln(x)ln(x)}{x^3}}dx = - \frac{1}{3}(ln(x))^3 +c_1$

$u_2 = \int{\frac{y_1(x)g(x)}{W[y_1, y_2](x)}}dx = \int{\frac{x^2 ln(x)}{x^3}}dx = \frac{(ln(x)^2)}{2} +c_2$

$y(x) = (- \frac{1}{3}(ln(x))^3 +c_1) x^2 + (\frac{(ln(x)^2)}{2}+c_2)x^2 ln(x) = \frac{1}{6}x^2(ln(x))^3 + c_1 x^2 + c_2 x^2 ln(x)$

The particular solution is thus

$y_p(x) = \frac{1}{6}x^2(ln(x))^3$
« Last Edit: October 27, 2018, 12:25:26 PM by Pengyun Li »

#### Victor Ivrii

You need to escape \ln etc so it will be typeset as $\ln x$