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Solve the equation: $$\quad y^{\prime \prime}+2 y\left(y^{\prime}\right)^{3}=0$$

$$x$$ is missing $$\rightarrow$$ Let $$y^{\prime}=u$$

$$y^{\prime \prime}=\frac{d u}{d x} \cdot \frac{d y}{d y}=\frac{d u}{d y} \cdot u$$

in $$D \cdot E :$$

$$\frac{d u}{d y} \cdot u+2 y(u)^{3}=0 \rightarrow 2 y(u)^{3}+\frac{d u}{d y} \cdot u=0 \quad * d y$$

$$2 y(u)^{3} d y+d u \cdot u=0 \quad \div u^{3} \rightarrow 2 y d y+\frac{d u}{u^{2}}=0$$

$$\frac{d u}{u^{2}}=u^{-2} d u$$



$$\frac{2 \cdot y^{2}}{2}-\frac{1}{u}=c \rightarrow$$ Where $$u=y^{\prime}=\frac{d y}{d x}$$

$$y^{2}-\frac{1}{d y / d x}=c \rightarrow y^{2}-\frac{d x}{d y}=c$$

$$\rightarrow y^{2}-c=\frac{d x}{d y} \quad * d y$$

$$\left(y^{2}-c\right) d y=\frac{d x}{d y} d y$$

$$\left(y^{2}-c\right) d y=d x$$

$$\rightarrow \frac{y^{3}}{3}-c_{1} y=x+c_{2} $$

Solve the equation:
$$y^{\prime \prime}=x+y^{2}, y(0)=1, y^{\prime}(0)=1$$

Non linear $$\rightarrow$$ Using Taylor method at 0:

$$y=y(0)+\frac{y^{\prime}(0) x}{1 !}+\frac{y^{\prime \prime}(0) x^{2}}{2 !}+\frac{y^{\prime \prime \prime}(0) x^{3}}{3 !}+\frac{y^{(4)}(0) x^{4}}{4 !} + ........$$

$$y^{\prime \prime}=x+y^{2} \rightarrow y^{\prime \prime}(0)=0+(1)^{2}=1$$
$$y^{\prime \prime \prime}=\left(x+y^{2}\right)^\prime=1+2 y y^{\prime}=1+2(1)(1)=3$$
$$y^{(4)}=\left(1+2 y y^{\prime}\right)=2 y^{\prime} y^{\prime}+2 y^{\prime \prime} y=2(1)(1)+2(1)(1)=2+2=4$$

$$y=1+x+\frac{1}{2} x^{2}+\frac{1}{2} x^{3}+\frac{1}{6} x^{4}+\cdots \cdots$$

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