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{\bf Question}

The differential equation

\[
y''+\delta(xy'+y)=0
\]

(here $\delta$ is a constant) arises in the study of turbulent
flow of a uniform stream past a circular cylinder. Verify that
$y_1(x)=\exp(-\delta x^2/2)$ is one solution and then find the
other solution in the form of an integral. $\qquad(*)$



\vspace{0.25in}

{\bf Answer}

Check that $y_1=e^x$ is a solution to the equation\\ This follows
since $y_1'=e^x$ and $y_1''=e^x$ and putting these into the
equation makes it true.\\ Equation is in standard form with $\ds
p(x)=\frac{-x}{x-1}$ and $\ds q(x)=\frac{1}{x-1}$. Use method of
reduction of order with $y_2=vy_1$ where $\ds v(x)=\int
\left(\frac{1}{y_1^2(x)}\;e^{-\int p(x) \;dx}\right)\;dx$\\ so
that $\ds v(x) =\int \left(\frac{1}{\left(e^x\right)^2} \;e^{-\int
\frac{-x}{x-1}\;dx} \right)\;dx$\\ $\Bigg\{$Note that $\ds
\int\frac{x}{x-1}\;dx=\int\frac{x-1+1}{x-1}\;dx =\int 1 +
\frac{1}{x-1}= x + \ln(x-1)\Bigg\}$\\ $\ds v(x)=\int
e^{-2x}\;e^{x+\ln(x-1)}\;dx =\int e^{-x}(x-1)\;dx =\int
xe^{-x}\;dx-\int e^{-x}\;dx$\\ $\ds v(x)=\int
xe^{-x}\;dx+e^{-x}=-xe^{-x}+\int e^{-x}\;dx+e^{-x}=-xe^{-x}$\\
Thus we have $\ds y_2(x)=\left(-xe^{-x}\right)e^x=-x$\\ Hence the
general solution is $\ds y(x)= Ae^x + Bx $


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