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


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

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

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

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

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

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

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

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

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

$$* 4$$

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


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

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

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

$$u=3 x+2 y$$

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

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

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

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

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

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

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

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

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

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

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

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

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

$$u=3 x+2 y$$

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

$$* 25$$

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

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

$$C_{1}=4 \ln 95$$

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

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