Toronto Math Forum
MAT2442018F => MAT244Tests => Term Test 1 => Topic started by: Victor Ivrii 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 ${y(0)=1, y'(0)=1}$.

(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:
$W(y_1,y_2)(x) = ce^{\lmoustachep(x)dx} = ce^{\lmoustache(2x/(x^{2} +1))dx} = c(x^2 + 1)$
(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:
$y_2 = y_1\lmoustache(e^{\lmoustachep(x)dx}/y_1^{2})dx = x\lmoustache(1 + 1/x^{2})dx = x(x  1/x) x^{2} 1$
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.

answer

(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
$$\begin{align}(x^{2}+1)\cdot 02x\cdot 1+2\cdot x&=0\\{2x+2x}&={0}\end{align}$$
$y_1(x)$ 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
$$y(x)=xx^{2}+1.$$

Yulin did everything right (but then why do you need to add scan? And please, no \lmoustache $\lmoustache$ next time, there is \int $\int$ !!)
Xiaoyuan, no point to post the scan after there is a perfect typed solution.
Monika, I know you mastered LaTeX :D, so no need to demonstrate your proficiency!