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Problems #14 from pg.40 and #2 on pg.48 are graded.
#14,pg.40: Solve the IVP
$$\left\{ \begin{array}{ll}
y'+2y=te^{-2t} \\
y(1)=0.
\end{array} \right.$$
Solution: This is a first order linear equation, so we will try using an integrating factor. It is already in standard form, so we see the integrating factor is
$$\mu(t)=\exp \left( \displaystyle\int 2 dt \right)=e^{2t}.$$
So multiply both sides by this and factor resulting in
$$(e^{2t}y)'=te^{-2t}e^{2t}=t.$$
Integration yields
$$e^{2t}y = \displaystyle\int t dt = \dfrac{t^2}{2}+C.$$
Therefore we have derived
$$y(t) = \dfrac{t^2}{2e^{2t}}+\dfrac{C}{e^{2t}}.$$
To apply the initial condition, compute
$$0=y(1)=\dfrac{1}{2e^2}+\dfrac{C}{e^2},$$
yielding $C=-\dfrac{1}{2}$. Therefore the solution of the IVP is
$$y(t)=\dfrac{t^2-1}{2e^{2t}}.$$
#2,pg.48: Solve $y'=\dfrac{x^2}{y(1+x^3)}$.
Solution: This ODE is separable, so that leads us to
$$(*) \hspace{35pt} \displaystyle\int y dy = \displaystyle\int \dfrac{x^2}{1+x^3} dx.$$
The left-hand-side yields $\dfrac{y^2}{2}$ while the right hand side can be integrated using the $u$-substitution $u=1+x^3$ so that $du = 3x^2 dx$, i.e. $\frac{1}{3}du=x^2 dx$ and so
$$\displaystyle\int \dfrac{x^2}{1+x^3} dx = \dfrac{1}{3} \displaystyle\int \dfrac{1}{u} du = \dfrac{1}{3} \log\left(|1+x^3|\right)+C$$
Thus $(*)$ becomes
$$\dfrac{y^2}{2} = \dfrac{1}{3} \log \left( | 1+x^3| \right)+C$$
Now solve for $y$ and we see the solution is
$$y(x) = \pm \sqrt{\dfrac{2}{3} \log(|1+x^3|)+C},$$
or written slightly differently:
$$y(x)=\pm \sqrt{\dfrac{2}{3}} \sqrt{\log(|1+x^3|)+C_1}; C_1=\frac{3}{2}C.$$
Note: in class we solved a similar problem, but decided on either the $+$ solution or the $-$ solution based on the initial condition -- this problem had no initial condition so we must keep both solutions.
