03 - Solving ODEs

Direct Integration (for First ODEs)

If the ODE takes the form of

$$\frac{dy}{dx}=f(x)$$

Then we can integrate both sides with respect to $x$ so

$$y=\int f(x)dx+c$$

Separation of Variables

If the ODE takes the form of

$$\frac{dy}{dx}=a(x)b(y)$$

where $a$ is a function of $x$ and $b$ is a function of $y$ then

$$\frac{1}{b(y)}\frac{dy}{dx}=a(x)$$

Hence by integrating with respect to $x$ then

$$\int \frac{1}{b(y)}dy=\int a(x)dx$$

(Assume that $b(y)\ne 0$ otherwise $y=c\in \mathbb{R}$ is a solution)

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Reduction to separable form by substitution

A reduction to separable form is a substitution that transforms a differential equation

$$\frac{dy}{dx}=F(x,y)$$

into an equation where variables can be separated, i.e.

$$g(y)\frac{dy}{dx}=f(x)$$

The goal is to find a substitution ( $u=\varphi (x,y)$ ) that simplifies the relation between ($x$) and ($y$) so that the resulting equation becomes separable.

Implicit Solutions

Solutions (typically to an ODE) where we have not found a $y$ in terms of $x$

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Solving First Order Linear Differential Equations

Suppose we have a first order differential equation ($k=1$)
Then we have the general form

$$\frac{dy}{dx}+p(x)y=q(x)$$

When $q(x)=0$, in other words in it’s homogenous form then it is separable

The inhomogeneous form can be solved using the integrating factor $I(x)$ by

$$I(x)=e^{\int p(x)dx}$$

Multiplying both sides by $I(x)$ then

$$e^{\int p(x)dx}\frac{dy}{dx}+p(x)e^{\int p(x)dx}y=e^{\int p(x)dx}q(x)$$

Using product rule then

$$\frac{d}{dx}(e^{\int p(x)dx}\cdot y)=e^{\int p(x)dx}q(x)$$

Hence we get

$$y=e^{-\int p(x)dx}\left[\int e^{\int p(x)dx}q(x)dx+c\right]$$

Solving Second Order Homogeneous Linear Differential Equations

Suppose $z(x)\ne 0$ then

$$p(x)\frac{d^{2}y}{d{x}^2}+q(x)\frac{dy}{dx}+r(x)y=0$$

Using substitution

$$y(x)=u(x)z(x)$$

such that

$$\frac{dy}{dx}=\frac{du}{dx}z+u\frac{dz}{dx}\text{ and }\frac{d^{2}y}{d{x}^2}=\frac{d^{2}u}{d{x}^2}+2\frac{du}{dx}\frac{dz}{dx}+u\frac{d^{2}z}{d{x}^2}$$

Then substituting these values into the ODE then

$$p(x)(u^{\prime \prime }z+2u^{\prime }{z}^{\prime }+u{z}^{\prime \prime })+q(x)(u^{\prime }z+u{z}^{\prime })+r(x)uz=0$$

As $z$ is a solution to the ODE then $p(x)z^{\prime \prime }+q(x)z^{\prime }+r(x)z=0$ hence

$$p(x)z{u}^{\prime \prime }+(2p(x)z^{\prime }+q(x)z)u^{\prime }=0$$

This is now a homogenous differential equation of first order for $u^{\prime }$ which should be solvable hence there is a general solution