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• Notes

$\frac{d y}{d x}=\tan ^{2}(x+y)$

$u=x+y$

$\frac{d u}{d x}=1+\frac{d y}{d x} \Rightarrow \frac{d y}{d x}=\frac{d u}{d x}-1$

$\frac{d u}{d x}-1=\tan ^{2} u$

$\frac{d u}{d x}=\tan ^{2} u+1$

$\frac{d u}{d x}=\sec ^{2} u=\frac{1}{\cos ^{2} u}$

$\cos ^{2} u d u=d x$

$\cos 2 u=2 \cos ^{2} u-1$

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

$\left[\frac{1}{2} \cos 2 u+\frac{1}{2}\right] d u=d x$

$\frac{1}{2} \frac{\sin 2 u}{2}+\frac{1}{2} u=x+C$

$* 4$

$\sin 2 u+2 u=4 x+4 c$

$u=x+y$

$\sin 2(x+y)+2 x+2 y=4 x+4 c$

$2 y+\sin [2(x+y)]=2 x+c_{1}$

$\frac{d y}{d x}=\frac{3 x+2 y}{3 x+2 y+2} \quad y(-1)=-1$

$u=3 x+2 y$

$\frac{d u}{d x}=3+2 \frac{d y}{d x}$

$\frac{d y}{d x}=\frac{1}{2}\left[\frac{d u}{d x}-3\right]$

$\frac{1}{2}\left[\frac{d u}{d x}-3\right]=\frac{u}{u+2}$

$\frac{d u}{d x}-3=\frac{2 u}{u+2}$

$\frac{d u}{d x}=3+\frac{2 u}{u+2}$

$\frac{d u}{d x}=\frac{3(u+2)+2 u}{(u+2)}$

$\frac{d u}{d x}=\frac{3 u+6+2 u}{u+2}$

$\frac{d u}{d x}=\frac{5 u+6}{u+2}$

$\frac{u+2}{5 u+6} d u=d x$

$\frac{u+2}{5 u+6}=\frac{1}{5}+\frac{4}{25 u+30}$

$\frac{1}{5} d u+\frac{4}{25 u+30} d u=d x$

$\frac{4}{25} \cdot \frac{4^{*} \frac{25}{4}}{25 u+30}=\frac{4}{25}\left[\frac{25}{25 u+30}\right]$

$\frac{1}{5} u+\frac{4}{25} \ln |25 u+30|=x+C$

$u=3 x+2 y$

$\frac{1}{5}(3 x+2 y)+\frac{4}{25} \ln |25(3 x+2 y)+30|=x+C$

$* 25$

$15 x+10 y+4 \ln |75 x+50 y+30|=25 x+25C$

$-15-10+4 \ln |-95|=-25+C_{1}$

$C_{1}=4 \ln 95$

$10 y-10 x+4 \ln |75 x+50 y+30|=4 \ln 95$