Differential Equations

First order differential equations: $\frac{{\rm d}y}{{\rm d}x}+Fy=G$​

$$
\begin{aligned}
&\frac{{\rm d}y}{{\rm d}x}+Fy=G\qquad z\frac{{\rm d}y}{{\rm d}x}+Fzy=Gz\qquad z\frac{{\rm d}y}{{\rm d}x}+\frac{{\rm d}z}{{\rm d}x}y=\frac{{\rm d}zy}{{\rm d}x}\\
&\frac{{\rm d}z}{{\rm d}x}=Fz\qquad\int F\text{ d}x=\int\frac{{\rm d}z}{z}=\ln z+c\qquad z=c\cdot e^{\int F\text{ d}x}\\
&\frac{{\rm d}zy}{{\rm d}x}=Gz\qquad zy=\int Gz\text{ d}x\qquad y=\frac{\int Gz\text{ d}x}{z}
\end{aligned}
$$

and find the particular solution from the initial conditions

Second order differential equations:

homogeneous differential equations: $a\frac{{\rm d}^2y}{{\rm d}x^2}+b\frac{{\rm d}y}{{\rm d}x}+cy=0$​

auxiliary equation: $\begin{align}
&\frac{{\rm d}y}{{\rm d}x}=\lambda y,\quad\frac{{\rm d}^2y}{{\rm d}x^2}=\frac{{\rm d}\frac{{\rm d}y}{{\rm d}x}}{{\rm d}x}=\frac{{\rm d}\lambda y}{{\rm d}x}=\lambda\frac{{\rm d}y}{{\rm d}x}=\lambda^2y\\
&\int\frac{y}{{\rm d}y}=\int\lambda\text{ d}x\qquad\ln y=\lambda x+c\qquad y=ce^{\lambda x}
\end{align}$

solve the equation $a\lambda^2+b\lambda+c=0$ for roots $\lambda_1,\lambda_2$ to find the complementary function:

1. if $\lambda_1=\lambda_2$, then $y=Ae^{\lambda_1x}+Be^{\lambda_2x}$
2. if $\lambda_1,\lambda_2$ are real, then $y=(Ax+B)e^{\lambda x}$
3. if $\lambda_1=m+ni,\lambda_2=m-ni$, then $y=e^{mx}(A\cos nx+B\sin nx)$​

find the particular solution from the boundary conditions

inhomogeneous differential equations: $a\frac{{\rm d}^2y}{{\rm d}x^2}+b\frac{{\rm d}y}{{\rm d}x}+cy=f(x)$

find the particular integral $g(x)$ such that $ag''(x)+bg'(x)+cg(x)=f(x)$

general solution = complementary function + particular integral

1. polynomial: $f(x)=f_tx^t+\cdots+f_1x+f_0$​

solve for the coefficients of $g(x)=g_tx^t+\cdots+g_1x+g_0$

2. exponential: $f(x)=pe^{qx}$​
$$
\begin{aligned}
&g(x)=\lambda e^{qx}\quad g'(x)=q\lambda e^{qx},g''(x)=q^2\lambda e^{qx}\\
&aq^2\lambda+bq\lambda+c\lambda=p\qquad\lambda=\frac{p}{aq^2+bq+c}
\end{aligned}
$$

3. trigonometric: $f(x)=p\sin kx+q\cos kx$
$$
\begin{aligned}
&g(x)=\alpha\sin kx+\beta\cos kx\\[4pt]
&g'(x)=k\alpha\cos kx-k\beta\sin kx,
g''(x)=-k^2\alpha\sin kx-k^2\beta\cos kx\\[4pt]
&\begin{cases}-ak^2\alpha-bk\beta+c\alpha=p\\-ak^2\beta+bk\alpha+c\beta=q\end{cases}
\Longrightarrow\begin{cases}(c-ak^2)\alpha-bk\beta=p\\bk\alpha+(c-ak^2)\beta=q\end{cases}
\end{aligned}
$$

Substitutions:

direct substitutions:

$$
z=\frac{ky}{x}:\\
z=px+qy:\\
z=xy:\\
y=zx:
$$

substitutions with the variable $t$:

$$
\begin{align}
&\frac{\text{d}y}{\text{d}x}=\frac{\text{d}y}{\text{d}t}\cdot\frac{\text{d}t}{\text{d}x}=\frac{\text{d}y}{\text{d}t}\cdot\frac{1}{x'(t)}=\frac{y'(t)}{x'(t)}\\[6pt]
&\frac{\text{d}^2y}{\text{d}x^2}=\frac{\text{d}\frac{\text{d}y}{\text{d}x}}{\text{d}x}=\frac{\text{d}\big(\frac{y'(t)}{x'(t)}\big)}{\text{d}t}\cdot\frac{\text{d}t}{\text{d}x}\\
=\,&\frac{x'(t)y''(t)-y'(t)x''(t)}{x'(t)\cdot x'(t)}\cdot\frac{1}{x'(t)}\\
=\,&\frac{y''(t)}{x'(t)\cdot x'(t)}-\frac{y'(t)\cdot x''(t)}{x'(t)\cdot x'(t)\cdot x'(t)}\\
=\,&\frac{\text{d}^2y}{\text{d}t^2}\cdot\frac{1}{x'(t)\cdot x'(t)}-\frac{\text{d}y}{\text{d}t}\cdot\frac{x''(t)}{x'(t)\cdot x'(t)\cdot x'(t)}
\end{align}
$$