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## First order linear ODEs

A first order linear ODE can be written as follows $\frac{dy}{dx}+p(x)y=g(x),$ where $$p$$ and $$g$$ are functions of $$x$$.

Steps to solve: Multiply both sides by the integrating factor $$\displaystyle e^{\int p(x)\, dx}$$: \begin{align*} \frac{dy}{dx}e^{\int p\, dx}+p(x)ye^{\int p\, dx}&=g(x)e^{\int p\, dx}\\ \frac{d}{dx}\left( ye^{\int p\, dx}\right)&=g(x)e^{\int p\, dx}\\ \int d\left( ye^{\int p\, dx}\right)&=\int \left( g(x)e^{\int p\, dx}\right) dx\\ ye^{\int p\, dx}&=\int \left( g(x)e^{\int p\, dx}\right) dx+c\\ y&=e^{-\int p\, dx}\left[\int \left( g(x)e^{\int p\, dx}\right) dx+c\right]\\ \end{align*}

Example. Solve the following first order linear ODE. $x\frac{dy}{dx}+(2x^2+1)y=2x,\; y(1)=2$ Find the valid interval of the solution. Find the behavior of $$y$$ when $$x\to \infty$$.

Solution. Let's write it in the standard form. $$$\frac{dy}{dx}+\left(2x+\frac{1}{x}\right)y=2,\; y(1)=2 \;\;\;\;(11)$$$ The IF is $$e^{\int\left(2x+\frac{1}{x}\right)\, dx}=e^{x^2+\ln|x|}=e^{x^2}e^{\ln|x|}=xe^{x^2}$$. Multiply both sides of (11) by the IF $$xe^{x^2}$$. \begin{align*} \frac{dy}{dx}xe^{x^2}+y\left(2x+\frac{1}{x}\right)xe^{x^2}&=2xe^{x^2}\\ \frac{d}{dx}\left(yxe^{x^2} \right)&=2xe^{x^2}\\ \int d\left(yxe^{x^2} \right)&=\int 2xe^{x^2}\, dx \text{ (substitute } u=x^2)\\ yxe^{x^2}&=e^{x^2}+c \end{align*} From the initial condition $$y(1)=2$$, we get $2\cdot 1e=e+c \implies c=e.$ Thus we get $$yxe^{x^2}=e^{x^2}+e$$ and the solution is $\displaystyle y=\frac{e^{x^2}+e}{xe^{x^2}}.$ The domain of $$y$$ is $$(-\infty,0)\cup (0,\infty)$$. Because of the initial condition $$y(1)=2$$, the valid interval of the solution is $$(0,\infty)$$. Note that $$\displaystyle y=\frac{1}{x}+\frac{e}{xe^{x^2}}\to 0$$ as $$x\to \infty$$.

Example. Solve the following first order linear ODE. $\frac{dy}{dt}+y\cos t=\cos t,\; y(0)=\pi$ Solution. The IF is $$e^{\int \cos t\, dt}=e^{\sin t}$$. Multiply both sides of the ODE by the IF $$e^{\sin t}$$. \begin{align*} \frac{dy}{dt}e^{\sin t}+y\cos t\; e^{\sin t}&=\cos t\; e^{\sin t}\\ \frac{d}{dt}\left(ye^{\sin t} \right)&=\cos t\; e^{\sin t}\\ \int d\left(ye^{\sin t} \right)&=\int \cos t\; e^{\sin t} \; dt\\ ye^{\sin t}&=\int e^u\, du && (u=\sin t,\, du=\cos t \, dt)\\ &=e^u+c && \\ &=e^{\sin t}+c && \end{align*} From the initial condition $$y(0)=\pi$$, we get $\pi\cdot e^0=e^0+c \implies c=\pi-1.$ Thus $$ye^{\sin t}=e^{\sin t}+\pi-1$$ and hence the solution is $\displaystyle y=\frac{e^{\sin t}+\pi-1}{e^{\sin t}}=1+(\pi-1) e^{-\sin t}.$

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