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Consider the equation $y^{\prime}-\frac{1}{3} y=\mathrm{e}^{-t}$
Find the solution with $y(0)=a$
Find the values of a for which (i) $y \rightarrow 0$ as $t \rightarrow \infty$ (ii) $y \rightarrow-\infty$ as $t \rightarrow \infty$

(a) $\frac{d y}{d t}-\frac{1}{3} y=e^{-t} \quad$ Linear in $y$

$\mu=e^{\int{\frac{-1}{3} d t}}=e^{-\frac{1}{3} t}$

The general solution: $\mu y=\int \mu Q(t) d t+c$

$e^{-\frac{1}{3} t} \cdot y=\int e^{-\frac{1}{3} t} \cdot e^{-t} d t+c$

$\Rightarrow e^{-\frac{1}{3} t} \cdot y=\int e^{-\frac{4}{3} t} d t+c$

$e^{-\frac{1}{3} t} \cdot y=-\frac{3}{4} \cdot e^{-\frac{4}{3} t}+c$

$e^{\frac{1}{3} t}\left(e^{-\frac{1}{3} t} \cdot y\right)=\left(-\frac{3}{4} \cdot e^{-\frac{4}{3} t}+c\right) e^{\frac{1}{3} t} \Rightarrow$

$y=\frac{-3}{4} \cdot e^{-t}+c\: e^{\frac{1}{3} t}$

$y(0)=a \Rightarrow$

$\frac{-3}{4} e^{0}+c\: e^{0}=-\frac{3}{4}+c=a$

$a=c-\frac{3}{4}$

$\Rightarrow c=a+\frac{3}{4}$

$y=\frac{-3}{4} e^{-t}+\left(a+\frac{3}{4}\right) e^{\frac{1}{3} t}$

$\lim _{t \rightarrow \infty} e^{t}=e^{\infty}=\infty$

$\lim _{t \rightarrow \infty} e^{-t}=e^{-\infty}=0$

$y=-\frac{3}{4} e^{-t}+\left(a+\frac{3}{4}\right) e^{1 / 3 t}$

(1) $\lim _{t \rightarrow \infty} y=0 \Rightarrow$

$-\frac{3}{4} e^{-\infty}+\left(a+\frac{3}{4}\right) e^{1 / 3(\infty)}=0$

$\Rightarrow a=-\frac{3}{4}$

(2) $\lim _{t \rightarrow \infty} y=-\infty \Rightarrow$

$-\frac{3}{4} e^{-\infty}+\left(a+\frac{3}{4}\right) e^{1 / 3(\infty)}=-\infty$

$a+\frac{3}{4}<0 \Rightarrow a<-\frac{3}{4}$

Find a solution for the equation $d x-(1+2 x \tan y) d y=0$

(1) $d x-(1+2 x \tan y) d y=0 \quad \div d y$

$\frac{d x}{d y}-(1+2 x \tan y)=0$

$\frac{d x}{d y}-1-2 x \tan y=0$

$\frac{d x}{d y}-(2 \tan y) x=+1$ linear in $x$

$\mu=e^{\int{-2} \tan y d y}=e^{-2 \ln |\sec y|}=\sec ^{-2} y$

$\int \tan (y)=\ln |\sec y|$

$a \ln |b|=\ln \left|b^{a}\right|$

$e^{\ln (x)}=x$

$\frac{1}{\sec y}=\cos y$

$=\frac{1}{\sec ^{2} y}=\cos ^{2}{y}$

The general solution:

$\mu x=\int \mu Q(y) d y+c$

$\frac{d x}{d y}-(2 \tan y) x=1$

$\cos ^{2} y \cdot x=\int \cos ^{2} y \cdot(1) d y+c$

$x \cos ^{2} y=\int\left(\frac{1}{2}+\frac{1}{2} \cos (2 y)\right) d y+c$

$\cos ^{2} y=\frac{1}{2}+\frac{1}{2} \cos 2 y$

$x \cos ^{2} y=\int \frac{1}{2}(1+\cos 2 y) d y+c \Rightarrow$

$x \cos ^{2} y=\frac{1}{2} \cdot\left(y+\frac{1}{2} \sin (2 y)\right)+c$