Question

. Solve the following initial value problem: (t2−20t+51)dydt=y (t2−20t+51)dydt=y with y(10)=1y(10)=1. (Find yy as a function of tt.)

Answer 1

Answer:

$y=(\frac{t-17}{t-3})^{\frac{1}{14}}$

Step-by-step explanation:

We are given that initial value problem

$t^2-20t+51)\frac{dy}{dt}=y$

$\frac{dy}{y}=\frac{dt}{t^2-20t+51}$

$\frac{dy}{y}=\frac{dt}{t^2-3t-17t+51}$

$\frac{dy}{y}=\frac{dt}{(t(t-3)-17(t-3)}$

$\frac{dy}{y}=\frac{dt}{(t-3)(t-17)}$

$\frac{1}{(t-3)(t-17)}=\frac{A}{t-3}+\frac{B}{t-17}$

$\frac{1}{(t-3)(t-17)}=\frac{A(t-17)+B(t-3)}{(t-3)(t-17)}$

$1=A(t-17)+B(t-3)$...(1)

Substitute t-3=0

t=3

t-17=0

t=17

Substitute t=3 in equation (1)

$1=A(3-17)+0$

$1=-14A$

$A=-\frac{1}{14}$

Substitute t=17

$1=B(17-3)$

$1=14B$

$B=\frac{1}{14}$

Substitute the values of A and B

$\frac{1}{(t-3)(t-17)}=-\frac{1}{14}(\frac{1}{t-3})+\frac{1}{14}(\frac{1}{t-17})$

$\int\frac{dy}{y}=-\frac{1}{14}\int\frac{dt}{t-3}+\frac{1}{14}\int\frac{dt}{t-17}$

$ln y=-\frac{1}{14}ln\mid{t-3}\mid+\frac{1}{14}\mid{t-17}\mid+ln C$

By using formula:$\frac{dx}{x}=ln x+C$

$ln y=\frac{1}{14}(-ln\mid{t-3}\mid+ln\mid{t-17}\mid)+ln C$

Using formula:$ln x-ln y=ln \frac{x}{y}$

$ln y=\frac{1}{14}(ln\mid{\frac{t-17}{t-3}}\mid)+ln C$

$ln y=\frac{1}{14}ln\mid{\frac{t-17}{t-3}}\mid+ ln C$

Substitute y(10)=1

$ln 1=\frac{1}{14}ln\mid\frac{10-17}{10-3}\mid+ln C$

$0=0+ln C$

Because ln 1=0

$lnC=0$

$C=e^0=1$

Because $ln x=y\implies x=e^y$

Substitute the value of C

$ln y=\frac{1}{14}ln\mid\frac{t-17}{t-3}\mid+ln1$

$ln y=\frac{1}{14}ln\mid\frac{t-17}{t-3}\mid+0$

$ln y=\frac{1}{14}ln\mid\frac{t-17}{t-3}\mid$

$14ln y=ln\mid\frac{t-17}{t-3}\mid$

$lny^{14}=ln\mid\frac{t-17}{t-3}\mid$

By using identity $blog a= loga^b$

$y^{14}=\frac{t-17}{t-3}$

$y=(\frac{t-17}{t-3})^{\frac{1}{14}}$