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Find a solution for the equation $$\frac{d y}{d x}=\frac{(1+2 x) e^{y}}{y+x^{2} y}$$

$$\frac{d y}{d x}=\frac{(1+2 x) e^{y}}{y+x^{2} y}$$

$$\left(y+x^{2} y\right) d y=(1+2 x) e^{y} d x$$

$$y\left(1+x^{2}\right) d y=(1+2 x) e^{y} d x \quad \div\left(1+x^{2}\right) e^{y}$$

$$y e^{-y} d y=\frac{1+2 x}{1+x^{2}} d x$$ Integrate

$$\int y e^{-y} d y=\int \frac{1+2 x}{1+x^{2}} d x$$

by parts

\begin{array}{ll}{u=y} & {d u=1 d y} \\ {v=-e^{-y}} & {d v=e^{-y}}\end{array}

$$-y e^{-y}-\int{-e}^{-y} d y=\int \frac{1}{1+x^{2}}+\frac{2 x}{1+ x^{2}} d x$$

$$\int \frac{1}{1+x^{2}}=\tan ^{-1} x$$

$$-y e^{-y}-e^{-y}=\tan ^{-1}(x)+\ln \left|1+x^{2}\right|+C$$

Find the general solution of the ODE $$\frac{d y}{d x}=\sin (x+y)$$

$$\frac{d y}{d x}=\sin (x+y) \quad \cdots \cdot(1)$$ put $$\quad t=x+y$$

$$\frac{d t}{d x}=1+\frac{d y}{d x} \rightarrow \frac{d y}{d x}=\frac{d t}{d x}-1 \cdots \cdot(2)$$

$$(2)=(1) \rightarrow \frac{d t}{d x}-1=\sin (t)$$

$$\frac{d t}{d x}=\sin (t)+1 \Rightarrow d t=(\sin (t)+1) d x \quad \div 1+\sin (t)$$

(1) $$\frac{d t}{1+\sin (t)}=d x$$

(2) $$\frac{d t}{1+\sin t}=d x$$ Integrate

$$\int \frac{d t \:(1-\sin(t))}{(1+\sin (t))(1-\sin (t))}=\int \frac{1-\sin (t)}{1-\sin ^{2}(t)} dt$$

$$=\int \frac{1-\sin (t)}{\cos ^{2} t} d t$$

$$\sin ^{2} x+\cos ^{2} x=1$$

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

$$=\int \frac{1}{\cos ^{2} t}-\frac{\sin t}{\cos ^{2} t} d t=\int \sec ^{2}(t)-\tan t \sec t d t$$

$$=\int \sec ^{2} t-\tan t \sec t d t=\tan t-\sec t$$

$$\tan t-\sec t=x+c$$

$$\Rightarrow \tan (x+y)-\sec (x+y)=x+c$$

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