Existence and Uniqueness theorem

Suppose p and q are functions continuous on $(a,b)$. Also assume $a<x_0<b$. Then the IVP $\frac{dy}{dx}+p(x)y=q(x),y(x_0)=y_0$ Has a unique solution on $(a,b)$

Example 2

Find the largest open interval we know there exists a unique solution $\frac{dy}{dx}+\frac{1}{x(x+2)}y=\frac{1}{x-5},y(3)=1$

Where are P and Q not continuous? Discontinuities at $x=[0,-2,5]$

Because the initial condition is at $x=3$, the interval must contain that point. By the existence uniqueness theorem , the interval of existence is $(0,5)$. We also know a unique solution exists by the EUT theorem on this interval.

Solving first order linear ODE

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

Multiply both sides by an integrating factor: $e^{\int p(x)dx}$

Reverse the product rule:

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

Example 1

Solve the IVP

$\frac{dy}{dx}+2xy=4x,y(0)=5$

Recall $\frac{dy}{dx}(y{e}^{\int 2xdx})=q(x)e^{\int 2xdx}$

$\frac{d}{dx}(y{e}^{x^2+c})=4x{e}^{x^2+c}$

$y{e}^{x^2}=\int 4x{e}^{x^2}dx$

$y{e}^{x^2}=\int 2e^{x^2}+c$

Real World Applications

Population

$\frac{dp}{dt}=kP$

$\ln (P)=kt+c$

$P(t)=e^{kt+c}$

$P(t)=P_0{e}^{kt}$

Example 1

Suppose that in two days a population of bacterial in a Petri dish has grown from 100,00o to 150,000. Estimate the bacteria population after 7 days.

$P=N_o{e}^{rt}$

$150,000=100,000e^{2r}$

$r=\frac{1}{2}\ln (\frac{3}{2})$

$P(7)=100,000e^{7\frac{1}{2}\ln (\frac{3}{2})}$

$=100,000({\frac{3}{2}}^{\frac{7}{2}})\approx 413,361$

Radioactive decay

$\frac{DQ}{dt}=-kQ,k>0$

Example 2

Suppose the quantity of a radioactive substance decreases from 50mg to 43mg after 5 days. How much will remain after 30 days.

$Q(t)=Q_o{e}^{-rt}$ where t is measured in days and $Q_0=Q_0=50mg$

$ans\approx 20.23mg$