### Author Topic: TT1 Problem 2 (noon)  (Read 2905 times)

#### Victor Ivrii

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##### TT1 Problem 2 (noon)
« on: October 16, 2018, 05:31:45 AM »
(a) Find Wronskian  $W(y_1,y_2)(x)$ of a fundamental set of solutions $y_1(x) , y_2(x)$ for ODE
\begin{equation*}
(x^2+1)y''-2xy'+2y=0.
\end{equation*}

(b) Check that $y_1(x)=x$ is a solution and find another linearly independent solution.

(c) Write the general solution. Find solution such that ${y(0)=1, y'(0)=1}$.

#### Yulin WANG

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• MAT244H1 2018F
##### Re: TT1 Problem 2 (noon)
« Reply #1 on: October 16, 2018, 09:46:21 AM »
(a) Rewrite the equation:

$y'' -(2x/x^{2} +1)y' + 2/(x^{2} + 1)y = 0$

Then p(x) = $-2x/(x^{2} +1)$

By Abel's Thereom, we have:

$W(y_1,y_2)(x) = ce^{\lmoustache-p(x)dx} = ce^{\lmoustache(2x/(x^{2} +1))dx} = c(x^2 + 1)$

(b) Since $y_1(x) = x$, so $y'_1(x) = 1 , y''_1(x) = 0$

Then plug in:

$0 - 2x/(x^{2} +1) + 2x/(x^{2} +1) = 0$

Thus, $y_1(x)$ is a solution.

Take c = 1, then $W(y_1,y_2)(x) = x^{2} + 1$

By Reduction of Oder, we can have:

$y_2 = y_1\lmoustache(e^{\lmoustache-p(x)dx}/y_1^{2})dx = x\lmoustache(1 + 1/x^{2})dx = x(x - 1/x) x^{2} -1$

Thus, $y_2(x) = x^{2} -1$

(c) By (b), we know $y = c_1x + c_2(x^{2} -1)$

Since y(0) = 1, y'(0) = 1

So $-c_2 = 1, c_1 = 1, then c_1 = 1, c_2 = -1$

Therefore, y = x- x^2 +1 is the solution to the IVP.
« Last Edit: October 18, 2018, 03:47:29 AM by Victor Ivrii »

#### Xiaoyuan Wang

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##### Re: TT1 Problem 2 (noon)
« Reply #2 on: October 16, 2018, 09:47:11 AM »

#### Monika Dydynski

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• Karma: 30
##### Re: TT1 Problem 2 (noon)
« Reply #3 on: October 16, 2018, 10:12:42 AM »
(a) Find Wronskian $W(y_1,y_2)(x)$ of a fundamental set of solutions $y_1(x)$, $y_2(x)$ for ODE

$$(x^{2}+1)y''-2xy'+2y=0.$$

Dividing but sides by $(x^{2}+1)$, we get
$$L[y]=y''-\frac{2x}{(x^{2}+1)}y'+\frac{2}{(x^{2}+1)}y=0,$$
where $p(x)=\frac{2x}{(x^{2}+1)}$, and $q(t)=\frac{2}{(x^{2}+1)}$.

By Abel's Theorem,

\begin{align}W(y_1,y_2)(x)&=c\exp(\int-{p(x)dx})\\&=c\exp(\int\frac{2x}{(x^{2}+1)}dx)\\&=ce^{\ln(x^{2}+1)}\\&=c(x^{2}+1).\end{align}

Let $c=1 \Rightarrow W(y_1,y_2)(x)=x^{2}+1$.

b) Check that $y_1(x)=x$ is a solution and find another linearly independent solution.

Since $y_1(x)=x \Rightarrow y_1 '(x)=1$, and $y_1 ''(x)=0$

Plugging $y_1$, $y_1 '$, and $y_1 ''$ into the ODE, we have

\begin{align}(x^{2}+1)\cdot 0-2x\cdot 1+2\cdot x&=0\\{-2x+2x}&={0}\end{align}
$y_1(x)$ satisfies the ODE $\Rightarrow$ $y_1(x)$ is a solution.

Given $y_1(x)$, we can find another linearly independent solution.

We know from the definition of the Wronskian that
$$W(y_1,y_2)(x)=y_1y_2 '-y_1 'y_2=xy_2 '-y_2$$

Equating the two expressions for the Wronskian, we get

$$xy_2 '-y_2=x^{2}+1$$

Dividing both sides by $x$, and multiplying by integrating factor $\mu=\frac{1}{x}$,

$$(\frac{1}{x}y_2)'=1+\frac{1}{x^2}$$
$$\frac{1}{x}y_2=\int{(1+\frac{1}{x^2})}dx+C$$
$$y_2(x)=x^{2}-1+Cx$$
$$y_2(x)=x^{2}-1$$

c)Write the general solution. Find solution such that $y(0)=1$, $y'(0)=1$

The general solution to the ODE is

$$y(x)=c_1 x+c_2(x^{2}-1).$$

$\Rightarrow y'(x)=c_1+2c_2 x$

$$1=c_1 \cdot 0+c_2(0^{2}-1)$$
$$1=c_1+2c_2 \cdot 0$$

$$\cases{c_1=1\\c_2=-1}$$

Thus, the solution that satisfies $y(0)=1$, $y'(0)=1$ is

$$y(x)=x-x^{2}+1.$$

« Last Edit: October 16, 2018, 10:26:19 AM by Monika Dydynski »

#### Victor Ivrii

Yulin did everything right (but then why do you need to add scan? And please, no \lmoustache $\lmoustache$ next time, there is \int $\int$ !!)