MAT244--2019F > Term Test 1

Problem 2 (morning)

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Victor Ivrii:
(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 y''-(2x+1)y'+(x+1)y=0.
\end{equation*}
(b) Check that $y_1(x)=e^x$ is a solution and find another linearly independent solution.

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

Jingjing Cui:
a)
$$y''-\frac{(2x+1)}{x}y'+\frac{x+1}{x}y=0\\ W=ce^{\int{-p(x)dx}}=ce^{\int{\frac{(2x+1)}{x}dx}}\\ \int{\frac{(2x+1)}{x}dx}=2x+ln|x|\\ W=ce^{2x+ln|x|}=cxe^{2x}\\$$

Jingjing Cui:
b)
$$y_1(x)=e^x\\ y_1'(x)=y_1''(x)=e^x\\ x(e^x)-(2x+1)e^x+(x+1)e^x=xe^x-2xe^x-e^x+xe^x+e^x=0\\ so \;\;y_1(x)=e^x \;is \;a \;solution\\ \\ take \;c=1, then \;W=xe^{2x}\\ W=det \begin{vmatrix} e^x&y_2(x)\\ e^x&y_2'(x)\\ \end{vmatrix}\\ e^xy_2'(x)-e^xy_2(x)=xe^{2x}\\ y_2'(x)-y_2(x)=xe^{x}\\ p(x)=-1\\ \mu=e^{\int{p(x)}dx}=e^{-x}\\ e^{-x}y_2'(x)-e^{-x}y_2(x)=x\\ e^{-x}y_2(x)=\int{x}dx=\frac{1}{2}x^2+c\\ y_2(x)=\frac{1}{2}x^2e^{x} \;(take\;c=0)\\$$

Jingjing Cui:
c)
$$y(x)=c_1(e^x)+c_2(\frac{1}{2}x^2e^x)\\ y(1)=c_1(e^1)+c_2(\frac{1}{2}1^2e^1)=c_1e+c_2(\frac{1}{2}e)=0\\ c_1=-\frac{1}{2}c_2\\ y'(x)=c_1(e^x)+c_2xe^x+c_2(\frac{1}{2}x^2e^x)\\ y'(1)=c_1(e^1)+c_2e^1+c_2(\frac{1}{2}1^2e^1)=c_1e+c_2e+\frac{1}{2}c_2e=e\\ c_1+c_2+\frac{1}{2}c_2=1\\ -\frac{1}{2}c_2+c_2+\frac{1}{2}c_2=1\\ c_2=1\\ c_1=-\frac{1}{2}\\ y(x)=-\frac{1}{2}e^x+\frac{1}{2}x^2e^x\\$$

Mengyuan Wang:
begin{equation}
\begin{array}{l}{y^{\prime \prime}-\frac{2 x+1}{x} y^{\prime}+(x+1) y=0} \\ {w=c e^{-f(x) d x}=C e^{b+\frac{1}{x} d x}=cx e^{2 x}}\end{array}

let $C=1$

w=\left|\begin{array}{cc}{e^{x}} & {y_{2}} \\ {e^{x}} & {y_{2}^{\prime}}\end{array}\right|=x e^{2 x}

\begin{array}{c}{e^{x} y^{\prime}-e^{x} y=x e^{2 x}} \\ {y^{\prime}-y=x e^{x}}\end{array}

u=e^{\int-1 d x}=e^{-x}

\begin{aligned} e^{-x} y^{\prime}-e^{-x} y &=x \\ e^{-x} y &=\int x d x \\ e^{-x} y &=\frac{x^{2}}{2}+c \end{aligned}

let $C=1$

y_{2}=\frac{x^{2} e^{x}}{2}+e^{x}

\item[c] so
\begin{array}{l}{y=c_{1} e^{x}+c_{2}\left(\frac{x^{2} e^{x}}{2}+e^{x}\right)} \\ {y^{\prime}=c_{1} e^{x}+c_{2}\left(\frac{2 x e^{x}}{2}+\frac{e^{x} x^{2}}{2}+e^{x}\right)}\end{array}

\begin{array}{l}{y(1)=0} \\ {y^{\prime}(1)=e}\end{array}

\begin{array}{l}{c_{1} e+G\left(\frac{e}{2}+e\right)=0} \\ {c_{1}+\frac{e}{2} c_{2}=0} \\ {c_{1} e+c_{2}\left(-e+\frac{e}{2}+e\right)= e} \\ {c_{1}+\frac{5}{2} c_{2}=1}\end{array}

\begin{array}{l}{C_{2}=1} \\ {C_{1}=-\frac{3}{2}}\end{array}

so

y=-\frac{3}{2} e^{x}+\left(\frac{x^{2}}{2} e^{-\frac{2}{2}}+e^{x}\right)

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