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{\bf Question}

Solve the following differential equations, using the specified
boundary conditions:
\begin{description}
\item[(a)]
$(x^2-3)\frac{dy}{dx}+2xy^2=0$,\ where $y=1$ when $x=2$.
(Separation of Variables)
\item[(b)]
$x\frac{dy}{dx}+y=\sin(x)$, where $y=1$ when $x=\frac{\pi}{2}$.
(Integrating Factor)
\item[(c)]
$\frac{dy}{dx}+2y=e^{3x}$, where $y=0$ when $x=0$. (Integrating
Factor)

\end{description}

{\bf Answer}

\begin{description}
\item[(a)]
\begin{eqnarray*}
(x^2-3) \frac{dy}{dx} & = & -2xy^2 \\ \int \frac{dy}{y^2} & = &
-\int \frac{2x}{x^2-3}\,dx + C \\ -\frac{1}{y} & = & -\ln |x^2-3|
+ C
\end{eqnarray*}
General solution: $\displaystyle y=\frac{1}{\ln |x^2-3|-C}$

\medskip Boundary condition: $y=1$ when $x=2$

$$1=\frac{1}{\ln (1)-C}=\frac{1}{-C}\Rightarrow C=-1$$

Particular solution: $\displaystyle y=\frac{1}{\ln |x^2-3|+1}$
\bigskip
\item[(b)]
Standard form: $\displaystyle \frac{dy}{dx}+\frac{y}{x}=
\frac{\sin x}{x}$

\medskip Integrating factor: $I(x)=e^{\int \frac{dx}{x}}=e^{\ln x}=x$

Multiply equation by integrating factor: $\displaystyle
x\frac{dy}{dx}+y=\sin x$

so \begin{eqnarray*} \frac{\partial{}}{\partial x}(xy) & = & \sin
x
\\ xy & = & \int \sin x\,dx + C
\end{eqnarray*}
\medskip General solution: $\displaystyle y=-\frac{\cos x}{x}+\frac{C}{x}$

Boundary condition: $y=1$ when $\displaystyle x=\frac{\pi}{2}$

$$1=-\frac{0}{\frac{\pi}{2}}+\frac{2C}{\pi} \Rightarrow
C=\frac{\pi}{2}$$

Particular solution: $\displaystyle y=-\frac{\cos x}{x} +
\frac{\pi}{2x}$
\bigskip
\item[(c)]
$\displaystyle \frac {dy}{dx}+2y=e^{3x}$

Integrating factor: $I(x)=e^{2\int\,dx}=e^{2x}$

Multiply equation by integrating factor:
\begin{eqnarray*}
e^{2x}\frac{dy}{dx}+2e^{2x}y & = & e^{5x}\\
\frac{\partial{}}{\partial x}(e^{2x}y) & = & e^{5x}
\end{eqnarray*}
so $\displaystyle e^{2x}y=\int e^{5x}\,dx + C$

General solution: $\displaystyle y=\frac{1}{5}e^{3x} + Ce^{-2x}$

Boundary condition: $y=0$\ when\ $x=0$

so $\displaystyle 0=\frac{1}{5}+C \Rightarrow C=-\frac{1}{5}$

Particular solution: $\displaystyle
y=\frac{1}{5}e^{3x}-\frac{1}{5}e^{-2x}$
\end{description}








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