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Separable ODEs


A separable ODE is of the form \[\frac{dy}{dx}=\frac{M(x)}{N(y)}.\] Steps to solve: \[\begin{align*} N(y)\,dy&=M(x)\, dx\\ \int N(y)\,dy&=\int M(x)\, dx + c.\\ \end{align*}\]

Example. Solve the following IVP (initial value problem) \begin{equation}\label{7} \frac{dy}{dx}=\frac{2x-3}{y-5},\; y(0)=3 \end{equation} and find the valid interval of the solution. Find the maximum value of the solution.

Solution. \begin{align*} \frac{dy}{dx}&=\frac{2x-3}{y-5} \\ (y-5)\,dy&=(2x-3)\,dx\\ \int (y-5)\,dy&=\int (2x-3)\,dx\\ \frac{y^2}{2}-5y&=2\frac{x^2}{2}-3x+c\\ y^2-10y&=2(x^2-3x+c)\\ y^2-10y-2x^2+6x-2c&=0 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; (8) \end{align*} Note that the initial condition is \(y(0)=3\). So we have \[\begin{align*} 3^2-10\cdot 3-2c&=0\\ 2c&=-21 \end{align*}\] From (8) we get \[\begin{align*} y^2-10y-2x^2+6x+21&=0 \text{ (implicit solution)}\\ y&=\frac{10\pm \sqrt{100-4(-2x^2+6x+21)}}{2}\\ y&=\frac{10\pm 2\sqrt{25-(-2x^2+6x+21)}}{2}\\ y&=5\pm \sqrt{2x^2-6x+4} \end{align*}\] So we have two possible solutions: \[y=5+ \sqrt{2x^2-6x+4} \text{ and } y=5- \sqrt{2x^2-6x+4}.\] Note that the first one does note satisfy the initial condition \(y(0)=3\). Thus the solution is \[y=5- \sqrt{2x^2-6x+4} \text{ (explicit solution)}.\] Obviously the maximum value of the solution is 5 because \(\sqrt{2x^2-6x+4}\geq 0\). The solution is defined when \[\begin{align*} 2x^2-6x+4&\geq 0\\ 2(x^2-3x+2)&\geq 0\\ 2(x-1)(x-2)&\geq 0 \end{align*}\] The domain of \(y\) is \((-\infty,1]\cup [2,\infty)\). Because of the initial condition \(y(0)=3\), the valid interval of the solution is \((-\infty,1]\).

Example. Solve the following ODE. \[\begin{equation} \frac{dy}{dx}=e^{-y}x\cos x \end{equation}\]

Solution. \[\begin{align*} \frac{dy}{dx}&=\frac{x\cos x}{e^y}&&\\ e^y\, dy&=x\cos x\, dx&\\ \int e^y\, dy&=\int x\cos x\, dx&\\ e^y &=\int u\, dv && u=x,\, dv=\cos x\, dx\\ &=uv-\int v\, du && du=dx,\, v=\int dv=\int \cos x\, dx=\sin x\\ &=x\sin x-\int \sin x\, dx &&\\ &=x\sin x+\cos x +c && \end{align*}\] The general solution is \(y=\ln|x\sin x+\cos x +c|\).

Substitution for homogeneous ODEs:
The ODE \(\displaystyle\frac{dy}{dx}=f(x,y)\) is called homogeneous if \(f(x,y)\) is a function of \(\displaystyle\frac{y}{x}\).

Steps to solve: Substitute \(v=y/x\) or, \(y=vx\). Then \(\displaystyle\frac{dy}{dx}=x\displaystyle\frac{dv}{dx}+v\). Write \(f(x,y)\) as a function of \(v\). Then it becomes a separable ODE of \(v\) and \(x\).

Example. Solve the following ODE. \[\begin{equation*} \frac{dy}{dx}=\frac{x^2+3y^2}{2xy} \end{equation*}\] Solution. \[\begin{align} \frac{dy}{dx}&=\frac{(x^2+3y^2)/x^2}{2xy/x^2}\nonumber\\ \frac{dy}{dx}&=\frac{1+3(y/x)^2}{2(y/x)} \;\;\;\;\; (9) \end{align}\] Let \(v=y/x\) or, \(y=vx\). Then \(\displaystyle\frac{dy}{dx}=x\displaystyle\frac{dv}{dx}+v\). So from (9) we get \[\begin{align} x\frac{dv}{dx}+v&=\frac{1+3v^2}{2v}\nonumber\\ x\frac{dv}{dx}&=\frac{1+3v^2}{2v}-v=\frac{1+v^2}{2v}\nonumber\\ \frac{2v}{1+v^2}\, dv&=\frac{dx}{x} \nonumber \end{align}\] \[\begin{align} \int\frac{2v}{1+v^2}\, dv&=\int \frac{dx}{x} \text{ (substitute } u=1+v^2) \nonumber\\ \ln|1+v^2|&=\ln|x|+\ln c \nonumber\\ \ln|1+v^2|&=\ln|cx| \nonumber\\ 1+v^2&=cx \nonumber\\ 1+\frac{y^2}{x^2}&=cx \nonumber\\ x^2+y^2&=cx^3 \nonumber \end{align}\]

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