Answer to Question #205359 in Differential Equations for Kate Bishop

Question

The rate of change of x is proportional to x. When t=0, x(0) =3, and when t=2, x=6. What js the value of x when t=4?

Answer 1

$Question :- \\ The \space rate \space of \space change \space of \space x \space is \space proportional \space to \space x.\\ When \\\space t=0, x(0) =3,\\ and \\when \\t=2, x(2)=6.\\ What \space the \space value \space of \space x \space when \space t=4? \\ -----------------------------------\\Solution:- \\given\\ The \space rate \space of \space change \space of \space x \space is \space proportional \space to \space x.\\ hence \\ \implies \frac{dx}{dt} \propto x\\ \implies \frac{dx}{dt} = k. x\\ \implies \frac{dx}{x} = k.dt\\ integration \space both \space side \\ \implies log(x)=k.t+ c\\ \implies x=e^{k.t+ c}\\ now \space \\ use \space first\space condition \\ When \\\space t=0, x(0) =3,\\ hence\\ \implies 3=e^{k.0+ c}\\ \implies 3=e^{c}\\ \implies c=log(3)\\ now \space \\ use \space second \space condition \\ When \\\space t=2, x(2) =6,\\ hence\\ \implies 6=e^{k.2 +log(3)}\\ \implies6=e^{k.2 }.3\\ \implies 2=e^{k.2}\\ \implies log(2)={k.2}\\ \implies 0.3010={k.2}\\ \implies k=0.1505\\ hence \\ \implies x=e^{k.t+ c}\\and \space k=0.1505, c=log(3)\\ hence\\ \implies x=e^{0.1505.t}.3\\ now \space we \space find \space the \space value \space of \space x \space when \space t=4?\\ \implies x=e^{0.1505.*4}.3\\ \implies x=5.477$