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: $$$2y\cdot y'=-x^2$$$ Now we transform $$y'=\dfrac{dy}{dx}$$

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

Solution:

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

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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 $$y\neq0$$ it is a separable ODE and we have: $$$y'=y\cdot\sin(x) \Rightarrow \dfrac{1}{y}\cdot\dfrac{y'}{y}=\sin(x) \Rightarrow \dfrac{1}{y}\cdot\dfrac{dy}{dx}=\sin(x) \Rightarrow$$$ $$$\Rightarrow \dfrac{dy}{y}=\sin(x)\cdot dx \Rightarrow \int \dfrac{dy}{y}=\int \sin(x)\cdot dx \Rightarrow$$$ $$$\ln|y(x)|=-\cos(x)+C \Rightarrow |y(x)|=e^{-\cos(x)+C}=e^{-\cos(x)}\cdot e^{C}=k\cdot e^{-\cos(x)}, \ k > 0 \Rightarrow$$$ $$$\Rightarrow y(x)=k\cdot e^{-\cos(x)}, \ k\neq0$$$ 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 $$k$$: $$$y(\pi)=-3 \Rightarrow -3=k\cdot e^{-\cos(\pi)}=k\cdot e \Rightarrow k=-\dfrac{3}{e}$$$

Solution:

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

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