MAT244--2019F > Term Test 1

Problem 2 (main)

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ZYR:
For question (b), I  think the first student who reply it forget to verify that $y_1(x) = xcos(x)$ is a solution.
Here is my solution: $y_1(x) = xcos(x)$
                              $y_1'(x) = cos(x) - xsin(x)$
                              $y_1''(x) = -sin(x) - sin(x) - xcos(x)$
Then substitute this to the original equation, we have
                              $(x^2y'' -2xy' + (x^2 + 2)y = 0$
$x^2( -sin(x) - sin(x) - xcos(x)) -2x(cos(x) - xsin(x)) +(x^2 +2） xcos(x)
= -2x^2sin(x) - x^3cos(x) -2xcos(x) +2x^2sin(x) + x^3cos(x) + 2xcos(x) = 0$

So,  $y_1(x) = xcos(x)$ is a solution.

Also, there is a mistake when the first student do the integral part for the question b), it should be $\int sec^2 x dx = tanx + c$

We are looking for another solution, not for all other solutions. No mistake. V.I.

Zhangxinbei:
For b)
I choose c = 1
= xcosx(tan x + 1)
= xcosxtanx + xcosx
= x cosx(six/cosx) + xcosx
= xsinx + xcosx

BJM:
Here is the solution.

Xinqiao Li:
a) Find Wronskian $W(y_1, y_2)(x)$ of a fundamental set of solutions $y_1(x)$, $y_2(x)$ for ODE $x^2y'' -2xy' + (x^2+2)y = 0$
b) Check that $y_1(x) = xcosx$ is a solution and find another linearly independent solution.
c) Write the general solution, and find solution that $y(\frac{\pi}{2}) = 1, y'(\frac{\pi}{2}) = 0$

a)
$$y'' - \frac{2}{x}y' + (1 + \frac{2}{x^2}) = 0$$
We see that $p(x) = -\frac{2}{x}$ is continuous everywhere except at $x=0$, $q(x) = (1 + \frac{2}{x^2})$ is continuous everywhere except at $x=0$.

Then by Abel's Theorem,
$$W(y_1, y_2)(x) = ce^{-\int p(x)dx} = ce^{\int(\frac{2}{x})dx} = ce^{2lnx} = cx^2$$
b)
Let's verify $y_1(x) = xcosx$ is a valid solution.

Since $y_1(x) = xcosx$, $y'_1(x) = cosx-xsinx$, $y''_1(x) = -sinx-xcosx-sinx = -xcosx-2sinx$

Plug in: $x^2y'' -2xy' + (x^2+2)y = 0 = -x^3cosx-2x^2sinx - 2xcosx + 2x^2sinx + x^3cosx + 2xcosx = 0$

So $y_1(x) = xcosx$ is a solution.

Take $c = 1, W(y_1, y_2)(x) = x^2$.

By Reduction of Order, we have:
$$y_2 = y_1\int(\frac{ce^{-\int p(x)dx}}{(y_1)^2})dx = xcosx\int(\frac{x^2}{(x^2cos^2x})dx = xcosx\int sec^2xdx = xcosx tanx = xsinx$$

c)
Then by part b, the general solution is $y = c_1xcosx +c_2xsinx$

The derivative is given by $y' = c_1(cosx -xsinx) + c_2(xcosx + sinx)$

Plug in the initial conditions $y(\frac{\pi}{2}) = 1, y'(\frac{\pi}{2}) = 0$, we have

$1 = \frac{\pi}{2}c_2$ and $0 = (-\frac{\pi}{2})c_1 + c_2$

So $c_2 = \frac{2}{\pi}$ and $c_1 = \frac{4}{(\pi)^2}$

The particular solution is,
$$y = \frac{4}{(\pi)^2}xcosx +\frac{2}{\pi}xsinx$$

yueyangyu:
a)
$$y'' - \frac{2}{x}y' + (1 + \frac{2}{x^2}) = 0$$

$$W(y_1, y_2)(x) = ce^{-\int p(x)dx} = ce^{\int(\frac{2}{x})dx} = ce^{2lnx} = cx^2$$

b)

Since $y_1(x) = xcosx$, $y'_1(x) = cosx-xsinx$, $y''_1(x) = -sinx-xcosx-sinx = -xcosx-2sinx$

 $x^2y'' -2xy' + (x^2+2)y = 0 = -x^3cosx-2x^2sinx - 2xcosx + 2x^2sinx + x^3cosx + 2xcosx = 0$

So $y_1(x) = xcosx$ is a solution.

Let $c = 1, W(y_1, y_2)(x) = x^2$.

$$y_2 = y_1\int(\frac{ce^{-\int p(x)dx}}{(y_1)^2})dx = xcosx\int(\frac{x^2}{(x^2cos^2x})dx = xcosx\int sec^2xdx = xcosx tanx = xsinx$$

c)
The general solution is $y = c_1xcosx +c_2xsinx$

 $y' = c_1(cosx -xsinx) + c_2(xcosx + sinx)$

 $y(\frac{\pi}{2}) = 1, y'(\frac{\pi}{2}) = 0$, we have

$1 = \frac{\pi}{2}c_2$ and $0 = (-\frac{\pi}{2})c_1 + c_2$

So $c_2 = \frac{2}{\pi}$ and $c_1 = \frac{4}{(\pi)^2}$

$$y = \frac{4}{(\pi)^2}xcosx +\frac{2}{\pi}xsinx$$

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