Problems from Separable ordinary differential equations

Development:

It is a separable ODE, since we can manage to separate the $x$'s and the $y$'s and put them on each side of the equation: $$2y\cdot y^{\prime }=-x^2$$ Now we transform $y^{\prime }={\displaystyle \frac{dy}{dx}}$

And we proceed as we have explained: $$2y\cdot {\displaystyle \frac{dy}{dx}}=-x^2\Rightarrow 2y\cdot dy=-x^2\cdot dx\Rightarrow \int 2y\cdot dy=\int -x^2\cdot dx\Rightarrow$$ $$\Rightarrow y^2=-{\displaystyle \frac{x^3}{3}}+C$$ Now, we solve for $y$ in terms of $x$: $$y(x)=\pm \sqrt{C-{\displaystyle \frac{x^3}{3}}},C\in \mathbb{R}$$ We observe that we have not obtained a unique solution. This is because $f(x,y)=-{\displaystyle \frac{x^2}{2y}}$ which is not continuous at $y=0$.

Solution:

$y(x)=\pm \sqrt{C-{\displaystyle \frac{x^3}{3}}},C\in \mathbb{R}$

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Development:

It is a separable ODE; we have to divide by $y$, so we will have to check if $y=0$ is also a solution. Therefore, we have to distinguish two cases:

Case 1: If $y\ne 0$ it is a separable ODE and we have: $$y^{\prime }=y\cdot \sin (x)\Rightarrow {\displaystyle \frac{1}{y}}\cdot {\displaystyle \frac{y^{\prime }}{y}}=\sin (x)\Rightarrow {\displaystyle \frac{1}{y}}\cdot {\displaystyle \frac{dy}{dx}}=\sin (x)\Rightarrow$$ $$\Rightarrow {\displaystyle \frac{dy}{y}}=\sin (x)\cdot dx\Rightarrow \int {\displaystyle \frac{dy}{y}}=\int \sin (x)\cdot dx\Rightarrow$$ $$\ln |y(x)|=-\cos (x)+C\Rightarrow |y(x)|=e^{-\cos (x)+C}=e^{-\cos (x)}\cdot e^C=k\cdot e^{-\cos (x)},k>0\Rightarrow$$ $$\Rightarrow y(x)=k\cdot e^{-\cos (x)},k\ne 0$$ where $k$ is a constant to be determined by the initial conditions.

Case 2: If $y=0$. We see that this is a solution. Therefore we also have to consider it, however it does not satisfy the initial condition

Thus the general solution is: $$y(x)=k\cdot e^{-\cos (x)},k\in \mathbb{R}$$ Now we can impose the initial conditions to determine $k$: $$y(\pi )=-3\Rightarrow -3=k\cdot e^{-\cos (\pi )}=k\cdot e\Rightarrow k=-{\displaystyle \frac{3}{e}}$$

Solution:

$y(x)=-3\cdot e^{-(cos(x)+1)}$

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