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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.